Proof. Let $U$ be a subspace of $X$. Let $p,q\in U$. Then there exist disjoint open subsets $P,Q$ of $X$ such that $p\in P$ and $q\in Q$. Note that $U\cap P$ and $Q\cap P$ are disjoint open subsets of $U$ such that $p\in U\cap P$ and $q\in U\cap Q$. Thus $U$ is Hausdorff. $\blacksquare$

Proof. Suppose for contradiction that $p\neq q$. Since $X$ is Hausdorff, there exists disjoint open subsets $A$ and $B$ such that $p\in A$ and $q\in B$. But by convergence, there exists $n\in N$ such that $p_n\in A$ and $p_n\in B$, implying $A\cap B\neq\emptyset$, a contradiction. Thus $p=q$. $\blacksquare$

Proof. Let $U\subseteq X$ be compact. If $U$ is empty, then it is trivially closed. Now suppose $U$ is non-empty. Let $p\in X\setminus U$. Then for all $q\in U$, there exist disjoint open sets $U_q,V_q$ such that $q\in U_q$ and $p\in V_q$. Note that $\{U_q:q\in U\}$ is an open cover of $U$ and thus has a finite subcover $\mathcal C$, and for some finite $S\subseteq U$, $\mathcal C=\{U_q:q\in S\}$. Since $U$ is non-empty, $S$ is non-empty. Then $\{V_q:q\in S\}$ is a non-empty finite set of open sets, thus $\cap\{V_q:q\in S\}$ is open. Note that $p\in\cap\{V_q:q\in S\}\subseteq X\setminus U$, implying $X\setminus U$ is a neighborhood of $p$. Since $X\setminus U$ is a neighborhood of every point in it, it is open. Thus $U$ is closed. $\blacksquare$

Proof. Suppose $X$ is locally compact. Let $p\in X$, then there exists a compact neighborhood $U_p$ of $p$, thus there exists an open set $S_p$ such that $p\in S_p\subseteq U_p$. Since $X$ is Hausdorff and $U_p$ is compact, $U_p$ is closed, implying $\overline{U_p}=U_p$. Let $\tau$ be the topology of $X$, and let $\mathcal B$ denote $\{S_p\cap U:p\in X,U\in\tau\}$. Note that given $p\in X$ and $U\in\tau$, $S_p\cap U$ is open and $\overline{S_p\cap U}\subseteq\overline{U_p}=U_p$, implying $\overline{S_p\cap U}$ is compact. Then every element of $\mathcal B$ is open and precompact. Let $S$ be an open subset of $X$. Then $S=\cup\{S_p\cap S:p\in X\}$ and $\{S_p\cap S:p\in X\}\subseteq\mathcal B$. Thus $\mathcal B$ is a precompact basis of $X$.

Suppose $X$ has a precompact basis $\mathcal B$. Then $X=\cup\mathcal B$. Let $p\in X$, then for some $U\in\mathcal B$, $p\in U$. Since $U$ is precompact, $\overline U$ is compact. Thus $\overline U$ is a compact neighborhood of $p$. Hence $X$ is locally compact. $\blacksquare$

Proof. Suppose $X$ is second-countable and let $(U_i)$ be a sequence of open sets as required by second-countability. Let $x\in X$, define $N_i=U_i$ if $x\in U_i$ and $N_i=X$ otherwise, then $(N_i)$ is a sequence of neighborhoods of $x$. Let $V$ be a neighborhood of $x$, then for some open set $S$, we have $x\in S\subseteq V$. Note that there exists $M\subseteq N$ such that $\cup\{U_i:i\in M\}=S$, thus for some $i\in M$, $x\in U_i$, implying $N_i=U_i\subseteq S\subseteq V$. We have shown that $X$ is first-countable. $\blacksquare$

Proof. Let $U$ be a subspace of $X$. Let $x\in U$. Then there exists a sequence $(N_i)$ of neighborhoods of $x$, with respect to $X$, such that every neighborhood $N$ of $x$, with respect to $X$, is a superset of $N_i$ for some $i\in N$. Then $(U\cap N_i)$ is a sequence of neighborhoods of $x$, with respect to $U$. Let $N$ be a neighborhood of $x$, with respect to $U$. Then $N\cup(X\setminus U)$ is a neighborhood of $x$, with respect to $X$, and thus is a superset of $N_i$ for some $i\in N$. Then $N$ is a superset of $U\cap N_i$. We have shown that $U$ is first-countable. $\blacksquare$

Proof. Let $(N_i)$ be a sequence of neighborhoods of $p$ as required by first-countability. Recursively define $M_0=N_0$ and $M_{i+1}=M_i\cap N_{i+1}$, then $(M_i)$ is a sequence of neighborhoods of $p$ satisfying the first-countability requirement, and $M_{i+1}\subseteq M_i$.

Suppose $p\in\overline S$. If $p\in S$, then a sequence $(p_i)$ such that $p_i=p$ trivially converges to $p$. Now suppose $p$ is a limit point of $S$. Then for each $i$, there exists $p_i\in S$ such that $p_i\in M_i\setminus\{p\}$. And $(p_i)$ forms a sequence of $S$. Let $U$ be a neighborhood of $p$, then for some $n\in N$, $M_n\subseteq U$, thus for all $k\ge n$, $M_k\subseteq U$, implying $p_k\in U$. Thus $(p_i)$ converges to $p$.

Suppose there exists a sequence $(p_i)$ of $S$ that converges to $p$. If $p\in S$, then trivially, $p\in\overline S$. Now suppose $p\notin S$. Let $U$ be a neighborhood of $p$, then for some $k\in N$, $p_k\in U$. Since $p_k\in S$, $p_k\neq p$. We have shown that $p$ is a limit point of $S$, and thus $p\in\overline S$. $\blacksquare$

Proof. Let $U$ be a subspace of $X$. Let $(U_i)$ be a sequence of open sets of $X$ as required by second-countability. Then $(U\cap U_i)$ is a sequence of open sets of $U$. Let $V$ be an open set of $U$, then for some open set $S$ of $X$, $V=S\cap U$. Then for some $M\subseteq N$, $S=\cup_{i\in M}\{U_i\}$. Then $V=\cup_{i\in M}\{U\cap U_i\}$. We have shown that $U$ is second-countable. $\blacksquare$

Proof. Let $(U_i)$ be a sequence of open sets as required by second-countability. Let $\mathcal C$ be an open cover of $X$. Define $M$ to be the set of natural numbers $k$ such that there exists $C\in\mathcal C$ with $U_k\subseteq C$. This defines a function $f:M\to\mathcal C$ such that for all $k\in M$, $U_k\subseteq f(k)$. Let $p\in X$, then for some $C\in\mathcal C$, $p\in C$. Since $C$ is open, for some $L\subseteq N$, $C=\cup\{U_i:i\in L\}$. Then for some $k\in L$, $p\in U_k\subseteq C$, implying $k\in M$. And we have $p\in U_k\subseteq f(k)$. Thus $\{f(k):k\in M\}$ covers $X$, which is a countable subcover of $\mathcal C$. $\blacksquare$

Proof. $R^0$ is trivially a $0$-manifold.

Let $n\gt0$, then $R^n$ is clearly a Hausdorff space and locally Euclidean of dimension $n$. The only thing to show is second-countability.

Let $X=\{B_r(p):r\in Q,p\in Q^n\}$, then $\abs{X}=\omega$, and there exists a tuple $(X_i)$ as a bijection from $\omega$ to $X$. Let $S$ be an open set of $R^n$ and let $p\in S$, then there exists $\delta\gt0$ such that $B_\delta(p)\subseteq S$. Note that there exists $q\in Q^n$ such that $0\lt d(p,q)\lt\frac{\delta}{3}$. Let $s\in Q$ such that $\frac{\delta}{3}\lt s\lt\frac{2\delta}{3}$, then $p\in B_s(q)\subseteq S$. Therefore, we can define a map $f:S\to X$ such that $p\in f(p)\subseteq S$ for all $p\in S$. And $S=\cup\{f(p):p\in S\}$. Note that $\{f(p):p\in S\}=\{X_i:i\in U\}$ for some $U\subseteq N$. Thus $\cup\{X_i:i\in U\}=S$.

We have shown that $R^n$ is an $n$-manifold. Then $H^n$, as a subspace of $R^n$, is Hausdorff and second-countable. Trivially, it is also locally half-Euclidean of dimension $n$. Thus $H^n$ is an $n$-manifold with boundary. $\blacksquare$

Proof. Let $T=\varphi(S)$. Then $\varphi|_S:S\to T$ is bijective, and $\varphi^{-1}(T)=S$. Since $\varphi$ is continuous, $\varphi|_S:S\to T$ is continuous. Since $\varphi^{-1}$ is continuous, $\varphi^{-1}|_T:T\to S$ is continuous. Since $\varphi^{-1}|_T=\varphi|_S^{-1}$, $\varphi|_S^{-1}$ is continuous. Thus $\varphi|_S:S\to T$ is a homeomorphism. Since $U$ is an open subset of $M$ and $S$ is an open subset of $U$, $S$ is an open subset of $M$. Since $\varphi^{-1}$ is continuous, the preimage of $\varphi^{-1}$ on $S$, which is $T$, is an open subset of $V$. Since $V$ is an open subset of $R^n$ (or $H^n$), $T$ is an open subset of $R^n$ (or $H^n$). Hence $\varphi|_S:S\to T$ is a coordinate chart on $M$. $\blacksquare$

Proof. Let $W$ be an open subset of $U$. We have shown that $\varphi_1|_W$ is indeed a chart on $M$. Since $W\cap V\subseteq U\cap V$, $\varphi_1(W\cap V)\subseteq\varphi_1(U\cap V)$ and $\varphi_2(W\cap V)\subseteq\varphi_2(U\cap V)$. Clearly, the transition map from $\varphi_1|_W$ to $\varphi_2$ and its inverse are just restrictions of the transition map from $\varphi_1$ to $\varphi_2$ and its inverse, and they are thus smooth. Hence $\varphi_1|_W$ and $\varphi_2$ are compatible. $\blacksquare$

Proof. Let $\mathcal A$ denote the smooth structure of $M$ and let $(\varphi,U)\in\mathcal A$. Since $(\varphi,U)$ is a chart, the restriction of $\varphi$ to an open subset $S$ of $U$, denoted $(\varphi,S)$, is also a chart. Let $(\psi,V)\in\mathcal A$, then $(\varphi,U)$ and $(\psi,V)$ are compatible, thus $(\varphi,S)$ and $(\psi,V)$ are compatible. Suppose for contradiction that $(\varphi,S)\notin\mathcal A$, then $\mathcal A\cup\{(\varphi,S)\}$ is a proper extension of $\mathcal A$ that is also a smooth atlas, implying $\mathcal A$ is not maximal, a contradiction. Thus $(\varphi,S)\in\mathcal A$. $\blacksquare$

Proof. The topological subspace $U$ is clearly Hausdorff and second-countable. Let $p\in U$, then there exists a chart $(\varphi,V)$ on $M$ such that $p\in V$, and $(\varphi,U\cap V)$ is a chart on $M$. Note that $U\cap V$ is an open subset of $U$, thus $(\varphi,U\cap V)$ is also a chart on $U$. We have shown that the topological subspace $U$ is an $n$-manifold (with boundary).

Let $\mathcal A$ denote the smooth structure of $M$ and let $\mathcal A'$ denote the set of restrictions of smooth charts of $M$ on $U$. For the same reason as above, every function in $\mathcal A'$ is a chart on $U$. Clearly, $\mathcal A'$ covers $U$ and is thus an atlas of $U$. Let $\varphi_1',\varphi_2'\in\mathcal A'$, then for some $\varphi_1,\varphi_2\in\mathcal A$, $\varphi_1',\varphi_2'$ are restrictions of $\varphi_1,\varphi_2$ on $U$. Since $\varphi_1$ and $\varphi_2$ are compatible, $\varphi_1'$ and $\varphi_2'$ are also compatible. Thus $\mathcal A'$ is a smooth atlas of $U$. Suppose some chart $\psi$ of $U$ is compatible with every chart in $\mathcal A'$ but not in $\mathcal A'$. Let $\varphi\in\mathcal A$, then the restriction $\varphi'$ of $\varphi$ on $U$ is in $\mathcal A'$, and thus compatible with $\psi$. Then $\varphi$ is also compatible with $\psi$. Thus $\mathcal A\cup\{\psi\}$ is a proper extension of $\mathcal A$ that is a smooth atlas of $M$, a contradiction. We have shown that $\mathcal A'$ is maximal. $\blacksquare$

Proof. Let $V_a$ denote the former and let $V_b$ denote the latter. A smooth manifold is a set with a topology and a smooth structure. Clearly, $V_a$ and $V_b$ share the same underlying set and the same topology. Let $\mathcal A$ denote the smooth structure of $M$. Then the smooth structure of $V_a$, denoted $\mathcal A_a$, is $\{\varphi|_V:\varphi\in\mathcal A\}$. And the smooth structure of $V_b$, denoted $\mathcal A_b$, is $\{\varphi|_V:\varphi\in\mathcal A_U\}$, where $\mathcal A_U$ denotes $\{\varphi|_U:\varphi\in\mathcal A\}$. Suppose $\psi\in\mathcal A_a$, then $\varphi|_V=\psi$ for some $\varphi\in\mathcal A$. Note that $\varphi|_U\in\mathcal A_U$, thus $\psi=\varphi|_V=(\varphi|_U)|_V\in\mathcal A_b$. Now suppose $\psi\in\mathcal A_b$, then for some $\phi\in\mathcal A_U$, $\phi|_V=\psi$, and for some $\varphi\in\mathcal A$, $\varphi|_U=\phi$. Thus $\psi=\phi|_V=(\varphi|_U)|_V=\varphi|_V$, implying $\psi\in\mathcal A_a$. We have shown that $V_a$ and $V_b$ also share the same smooth structure. $\blacksquare$

Proof. Since $\mathcal A\subseteq\overline{\mathcal A}$, $\overline{\mathcal A}$ is an atlas of $M$. Let $\varphi_1:U_1\to V_1,\varphi_2:U_2\to V_2$ be charts in $\overline{\mathcal A}$. Let $x\in\varphi_1(U_1\cap U_2)$. Since $\mathcal A$ is an atlas of $M$, there exists $\varphi_3:U_3\to V_3$ in $\mathcal A$ such that $\varphi_1^{-1}(x)\in U_3$. Since $\varphi_1$ and $\varphi_3$ are compatible, $\varphi_3\circ\varphi_1^{-1}$ is diffeomorphic. Since $\varphi_2$ and $\varphi_3$ are compatible, $\varphi_2\circ\varphi_3^{-1}$ is diffeomorphic. Since $\varphi_1^{-1}(x)\in U_1\cap U_2\cap U_3$, $x\in\varphi_1(U_1\cap U_2\cap U_3)$, which is open. Note that $(\varphi_2\circ\varphi_3^{-1})\circ(\varphi_3\circ\varphi_1^{-1})$, when restricted from $\varphi_1(U_1\cap U_2\cap U_3)$ to $\varphi_2(U_1\cap U_2\cap U_3)$, is a diffeomorphism. Let $u\in\varphi_1(U_1\cap U_2\cap U_3)$, then $(\varphi_2\circ\varphi_3^{-1})\circ(\varphi_3\circ\varphi_1^{-1})(u)=\varphi_2\circ\varphi_1^{-1}(u)$. If $\varphi_1(U_1\cap U_2\cap U_3)$ is an open subset of $H^n$, then we need an extra step to find a smooth map on an open ball centered at $x$ that agrees with $\varphi_2\circ\varphi_1^{-1}$ on its domain if $x_n\gt0$ or on its intersection with $H^n$ if $x_n=0$. We have shown that $\varphi_2\circ\varphi_1^{-1}$ is smooth. By a symmetric argument, $\varphi_1\circ\varphi_2^{-1}$ is also smooth. Thus $\varphi_2\circ\varphi_1^{-1}$ is diffeomorphic, implying $\varphi_1$ and $\varphi_2$ are compatible. Hence $\overline{\mathcal A}$ is a smooth atlas of $M$.

Now suppose for contradiction that there exists a proper extension $\overline{\mathcal A}'$ of $\overline{\mathcal A}$ that is a smooth atlas. Then some chart $\varphi$ is in $\overline{\mathcal A}'$ but not in $\overline{\mathcal A}$. And $\varphi$ is compatible with every chart in $\overline{\mathcal A}'$, and thus every chart in $\mathcal A$. But then $\varphi$ should be in $\overline{\mathcal A}$, a contradiction. Hence $\overline{\mathcal A}$ is maximal. $\blacksquare$

Proof. For $i\in\{1,\ldots,n+1\}$, define $g_i^+:B(n,r,\hat p_i)\to U_i^+$ such that $g_i^+(y)=(y_1,\ldots,y_{i-1},p_i+\sqrt{r^2-d(y,\hat p_i)^2},y_i,\ldots,y_n)$ for all $y\in B(n,r,\hat p_i)$ and $g_i^-:B(n,r,\hat p_i)\to U_i^-$ such that $g_i^-(y)=(y_1,\ldots,y_{i-1},p_i-\sqrt{r^2-d(y,\hat p_i)^2},y_i,\ldots,y_n)$ for all $y\in B(n,r,\hat p_i)$. Then for each $i$, for all $x\in U_i^+$, $g_i^+(f_i^+(x))=x$, for all $x\in U_i^-$, $g_i^-(f_i^-(x))=x$, and for all $y\in B(n,r,\hat p_i)$, $f_i^+(g_i^+(y))=y$ and $f_i^-(g_i^-(y))=y$. Thus $f_i^+$ and $f_i^-$ are bijective, and $g_i^+$ and $g_i^-$ are their inverses respectively. Since these functions are trivially continuous, they are homeomorphisms.

Let $i,j\in\{1,\ldots,n+1\}$.

• If $i=j$, then $f_j^+\circ g_i^+$ is an identity map, which is smooth, and the domain of $f_j^-\circ g_i^+$ is empty, thus it is smooth.
• If $i\neq j$, then the component functions of $f_j^+\circ g_i^+$ are either a projection function or $p_i+\sqrt{r^2-d(y,\hat p_i)^2}$, thus $f_j^+\circ g_i^+$ is smooth, and similarly, $f_j^-\circ g_i^+$ is also smooth.

By a symmetric argument, transition maps from $g_i^-$ to $f_j^+$ or $f_j^-$ are smooth. $\blacksquare$

Proof. Let $(\varphi,U)$ be a smooth chart of $M$, then $\varphi(U)$ is an open subset of either $R^n$ or $H^n$. Suppose $\varphi(U)$ is both an open subset of $R^n$ and an open subset of $H^n$ such that $\varphi(U)\cap\partial H^n\neq\emptyset$, then there exists $p\in\varphi(U)$ such that $p\in\partial H^n$. But then no open ball centered at $p$ is contained by $\varphi(U)$, contradicting that $\varphi(U)$ is an open subset of $R^n$. Hence $\varphi$ cannot be both an interior chart and a boundary chart. If $\varphi(U)$ is an open subset of $R^n$, then it is an interior chart. If $\varphi(U)$ is an open subset of $H^n$, then

• if $\varphi(U)\cap\partial H^n=\emptyset$, then $\varphi(U)$ is also an open subset of $R^n$, thus it is an interior chart;
• if $\varphi(U)\cap\partial H^n\neq\emptyset$, then it is a boundary chart.

$\blacksquare$

Proof. Let $y\in f(U)$, then there exists $x\in U$ such that $f(x)=y$. By inverse function theorem, there exists an open subset $V$ of $U$ containing $x$ such that $f(V)$ is open. Since $y=f(x)\in f(V)$, there exists an open ball $B$ centered at $y$ such that $B\subseteq f(V)\subseteq f(U)$. Therefore $f(U)$ is open. $\blacksquare$

Proof. Suppose for contradiction that $U$ and $V$ are diffeomorphic, then there exists a diffeomorphism $f:U\to V$. Since $V\cap\partial H^n\neq\emptyset$, there exists $y\in V\cap\partial H^n$. Let $x$ denote $f^{-1}(y)$. Then there exists an open subset $W$ of $R^n$ containing $y$ and a smooth function $g:W\to R^n$ that agrees with $f^{-1}$ on $W\cap V$. And we can find an open ball $B$ centered at $x$ such that $B\subseteq U$ and $f(B)\subseteq W\cap V$. Then $g\circ f|_B$ is the identity function on $B$. By chain rule, for every $u\in B$, $I_n=D(g\circ f|_B)(u)=Dg(f|_B(u))D(f|_B)(u)$, implying $D(f|_B)(u)$ is invertible. Therefore, $f(B)$ is an open subset of $R^n$ by the above lemma. Since $y=f(x)\in f(B)\subseteq V\subseteq H^n$, there exists an open ball centered at $y$ that is a subset of $H^n$. But we also have $y\in\partial H^n$, a contradiction. $\blacksquare$

Proof. Let $M$ be a smooth $n$-manifold with boundary. If $n=0$, then every point of $M$ is an interior point and not a boundary point. Now suppose $n\neq0$.

Let $p\in M$. Then some smooth chart $(\varphi,U)$ covers $p$.

• If $\varphi(U)$ is an open subset of $R^n$, then $p$ is an interior point.
• If $\varphi(U)$ is an open subset of $H^n$:
  • if $\varphi(p)\in\partial H^n$, then $p$ is a boundary point,
  • if $\varphi(p)\in\Int H^n$, let $B$ be an open ball centered at $\varphi(p)$ covered by the codomain of $\varphi$, then the preimage $A$ of $B$ is an open subset of $M$, and $\varphi|_A:A\to B$ is a smooth chart that covers $p$ with the codomain $B$ being an open subset of $R^n$, hence $p$ is an interior point.

We have shown that $p$ is either an interior point or a boundary point.

Suppose for contradiction that $p\in M$ is both an interior point and a boundary point. Then there exist a smooth chart $(\varphi_1,U_1)$ covering $p$ such that $\varphi_1(U_1)$ is an open subset of $R^n$ and a smooth chart $(\varphi_2,U_2)$ covering $p$ such that $\varphi_2(U_2)$ is an open subset of $H^n$ and $\varphi_2(p)\in\partial H^n$. Since $\varphi_1$ and $\varphi_2$ are compatible, $\varphi_2\circ\varphi_1^{-1}:\varphi_1(U_1\cap U_2)\to\varphi_2(U_1\cap U_2)$ is a diffeomorphism. Since $U_1\cap U_2$ is an open subset of both $U_1$ and $U_2$, $\varphi_1(U_1\cap U_2)$ is an open subset of $\varphi_1(U_1)$ and $\varphi_2(U_1\cap U_2)$ is an open subset of $\varphi_2(U_2)$. Thus $\varphi_1(U_1\cap U_2)$ is an open subset of $R^n$ and $\varphi_2(U_1\cap U_2)$ is an open subset of $H^n$. Since $p\in U_1\cap U_2$, $\varphi_2(p)\in\varphi_2(U_1\cap U_2)$, and since we also have $\varphi_2(p)\in\partial H^n$, we have $\varphi_2(U_1\cap U_2)\cap\partial H^n\neq\emptyset$. Then by the above lemma, $\varphi_1(U_1\cap U_2)$ and $\varphi_2(U_1\cap U_2)$ are not diffeomorphic, a contradiction. $\blacksquare$

Proof. Let $p\in M$, then there exists a smooth chart $(\varphi,U)$ of $M$ such that $p\in U$. Since $\varphi(p)\in\varphi(U)$, $\varphi(U)$ is non-empty and equals $R^0$, implying $p$ is an interior point. By injectivity of $\varphi$, $U=\{p\}$. Since any smooth chart $(\varphi,U)$ of $M$ such that $p\in U$ would satisfy these conditions, such chart is unique.

Since $M$ is second-countable, there exists a countable basis $B$ of $M$. Let $p\in M$, then $\{p\}$ is open, thus there exists a subset $S$ of $B$ such that $\{p\}=\cup S$. For all $s\in S$, $s\subseteq\{p\}$, thus $s$ is either $\emptyset$ or $\{p\}$. Since not all $s\in S$ is empty, some $s\in S$ is $\{p\}$, thus $\{p\}\in B$. Thus $$\abs{M}=\abs{\{\{p\}:p\in M\}}\le\abs{B}\le\omega$$ $\blacksquare$

Proof. Let $M$ be a smooth $n$-manifold, then it is an $n$-manifold with boundary. The smooth structure $\mathcal A$ of $M$ is clearly a smooth atlas of $M$ as a manifold with boundary. Now suppose for contradiction that we have a proper extension $\mathcal A'$ of $\mathcal A$ that is a smooth atlas of $M$ as a manifold with boundary. If some chart $(\varphi,U)$ in $\mathcal A'$ has $\varphi(U)$ being an open subset of $H^n$ such that $\varphi(U)\cap\partial H^n\neq\emptyset$, then there exists $p\in U$ such that $\varphi(p)\in\partial H^n$, implying $n\neq0$. Note that for some chart $(\psi,V)$ in $\mathcal A$ and thus in $\mathcal A'$, $p\in V$ and $\psi(V)$ is an open subset of $R^n$. Then $\psi\circ\varphi^{-1}$ is a diffeomorphism, a contradiction to a lemma above. Thus every chart $(\varphi,U)$ in $\mathcal A'$ has $\varphi(U)$ being an open subset of $R^n$, implying $\mathcal A'$ is a smooth atlas of $M$ as a manifold without boundary, contradicting to $\mathcal A$ being maximal. We have shown that $\mathcal A$ is a smooth structure of $M$ as a manifold with boundary. Hence $M$ is a smooth $n$-manifold with boundary. $\blacksquare$

Proof. Let $p\in\Int M$, then for some interior chart $(\varphi,U_p)$, $p\in U_p$. For every $u\in U_p$, we can find an open ball $B_u\subseteq\varphi(U_p)$ centered at $\varphi(u)$, and $(\varphi,\varphi^{-1}(B_u))$ is an interior chart and $u\in\varphi^{-1}(B_u)$, implying $u\in\Int M$. Thus $U_p\subseteq\Int M$. Hence $\Int M=\bigcup_{p\in\Int M}U_p$ is an open subset of $M$. Then $\Int M$ is an open submanifold with boundary of $M$. Clearly, every point of $\Int M$ is locally Euclidean, thus $\Int M$ is an $n$-manifold. Suppose for contradiction that some smooth chart $(\varphi,U)$ of $\Int M$ is a boundary chart. Then for some $p\in U$, $\varphi(p)\in\partial H^n$, thus $p$ is a boundary point of $\Int M$. But $p$ is an interior point of $M$ and thus an interior point of $\Int M$ as shown above, a contradiction. Thus every smooth chart of $\Int M$ is an interior chart, and the smooth structure of $\Int M$ as a manifold with boundary is a smooth structure of $\Int M$ as a manifold. Hence $\Int M$ is a smooth $n$-manifold, without boundary.

Suppose $n\neq0$. $\partial M$ is clearly Hausdorff and second-countable. Let $p\in\partial M$, then there exists a boundary chart $(\varphi,U)$ such that $p\in U$ and $\varphi(p)\in\partial H^n$. Let $S$ denote $\varphi(U)\cap\partial H^n$ and define $f:R^n\to R^{n-1}$ by $f(x_1,\ldots,x_n)=(x_1,\ldots,x_{n-1})$, then $f|_S:S\to f(S)$ is a diffeomorphism and thus a homeomorphism. Let $y\in f(S)$, then $(f|_S)^{-1}(y)=(y_1,\dots,y_{n-1},0)\in S$, which we will denote as $x$. Note that for some open subset $V$ of $R^n$, $V\cap H^n=\varphi(U)$. And we have $x\in V$. Then for some $r\gt0$, $B_r(x)\subseteq V$. Let $y'\in B_r(y)$, and denote $(f|_S)^{-1}(y')=(y'_1,\dots,y'_{n-1},0)$ as $x'$, then $x'\in B_r(x)$, thus $x'\in V\cap\partial H^n$, implying $x'\in S$. Hence $y'=(f|_S)(x')\in f(S)$. We have shown that $f(S)$ is an open subset of $R^{n-1}$. Note that $\varphi^{-1}|_S:S\to\varphi^{-1}(S)$ is homeomorphic and $\varphi(p)\in S$, thus $p\in\varphi^{-1}(S)$. Let $q\in U\cap\partial M$. If $\varphi(q)\in\Int H^n$, then $q$ is an interior point, a contradiction. Thus $\varphi(q)\in S$ and $q\in\varphi^{-1}(S)$. Let $q\in\varphi^{-1}(S)$, then $q\in U$ and $\varphi(q)\in\partial H^n$, implying $q\in U\cap\partial M$. We have shown that $\varphi^{-1}(S)=U\cap\partial M$, which is an open subset of $\partial M$. Hence $f|_S\circ\varphi|_{U\cap\partial M}:U\cap\partial M\to f(S)$ is a homeomorphism from an open subset of $\partial M$ containing $p$ to an open subset of $R^{n-1}$. Therefore, $\partial M$ is an $(n-1)$-manifold. Similarly, we can see that for every boundary chart $(\varphi,U)$ of $M$, $f\circ\varphi|_{U\cap\partial M}$ (with codomain being the range) is a chart of $\partial M$, and these charts form a smooth atlas, which generates a smooth structure of $\partial M$.

If $M$ is a smooth $n$-manifold, then every point of $M$ is an interior point, thus $\partial M=\emptyset$. If $\partial M=\emptyset$, then $M=\Int M$, which is a smooth $n$-manifold. $\blacksquare$

Proof. Let $p\in M$, then there exists a smooth chart $(\varphi_f,U_f)$ where $p\in U_f$ such that $f\circ\varphi_f^{-1}$ is smooth, and there exists a smooth chart $(\varphi_g,U_g)$ where $p\in U_g$ such that $g\circ\varphi_g^{-1}$ is smooth. Note that $U_f\cap U_g$ is an open subset of both $U_f$ and $U_g$ and $p\in U_f\cap U_g$. Thus $(\varphi_f,U_f\cap U_g)$ and $(\varphi_g,U_f\cap U_g)$ are smooth charts of $M$, implying $\varphi_g\circ\varphi_f^{-1}$ is smooth. Hence $g\circ\varphi_f^{-1}=(g\circ\varphi_g^{-1})\circ(\varphi_g\circ\varphi_f^{-1})$ is smooth. $\blacksquare$

Proof. Let $p\in M$, then there exists a smooth chart $(\varphi,U)$ where $p\in U$ such that both $f\circ\varphi^{-1}$ and $g\circ\varphi^{-1}$ are smooth. Thus $(f+g)\circ\varphi^{-1}=(f\circ\varphi^{-1})+(g\circ\varphi^{-1})$ is smooth. Hence $f+g$ is smooth.

Let $p\in M$, then there exists a smooth chart $(\varphi,U)$ where $p\in U$ such that both $f\circ\varphi^{-1}$ and $c\circ\varphi^{-1}$ are smooth. Thus $(cf)\circ\varphi^{-1}=(c\circ\varphi^{-1})(f\circ\varphi^{-1})$ is smooth. Hence $cf$ is smooth. $\blacksquare$

Proof. Let $M,N$ be smooth manifolds with boundary. Let $f:M\to N$ be a smooth map. Let $B$ be an open subset of $N$ and let $A$ be the preimage of $f$ on $B$. Let $p\in A$, then there exist a smooth chart $(\varphi_M,U)$ where $p\in U$ and a smooth chart $(\varphi_N,V)$ where $f(p)\in V$ such that $f(U)\subseteq V$ and $\varphi_N\circ f\circ\varphi_M^{-1}:\varphi_M(U)\to\varphi_N(V)$ is smooth, and thus continuous. Then $f|_U=\varphi_N^{-1}\circ(\varphi_N\circ f\circ\varphi_M^{-1})\circ\varphi_M$, which is continuous. Note that $V\cap B$ is an open subset of $N$, and $U\cap A$ is the preimage of $f|_U$ on $V\cap B$ and is thus open with respect to $U$ and thus $M$. We have shown that for all $p\in A$, there exists an open subset $U_p$ of $M$ such that $p\in U_p\subseteq A$. Note that $\cup\{U_p:p\in A\}=A$, hence $A$ is open. And we have shown that $f$ is continuous. $\blacksquare$

Proof. Suppose $f$ is smooth and $U$ is an open submanifold of $M$. Let $p\in U$, then there exist a smooth chart $(\varphi_M,V)$ where $p\in V$ and a smooth chart $(\varphi_N,W)$ where $f(p)\in W$ such that $f(V)\subseteq W$ and $\varphi_N\circ f\circ\varphi_M^{-1}:\varphi_M(V)\to\varphi_N(W)$ is smooth. Note that $(\varphi_M,U\cap V)$ is a smooth chart on $U$, and $\varphi_N\circ f\circ\varphi_M^{-1}:\varphi_M(U\cap V)\to\varphi_N(W)$ is a restriction of $\varphi_N\circ f\circ\varphi_M^{-1}:\varphi_M(V)\to\varphi_N(W)$ and is thus smooth. Thus the restriction of $f$ on $U$ is smooth.

Suppose for each $p\in M$, there exists an open submanifold $U$ of $M$ where $p\in U$ such that $f|_U$ is smooth. Let $p\in M$, then there exists an open submanifold $U$ of $M$ where $p\in U$ such that there exist a smooth chart $(\varphi_U,V)$ of $U$ where $p\in V$ and a smooth chart $(\varphi_N,W)$ where $f(p)\in W$ such that $f(V)\subseteq W$ and $\varphi_N\circ f\circ\varphi_U^{-1}:\varphi_U(V)\to\varphi_N(W)$ is smooth. Note that $(\varphi_U,V)$ is also a smooth chart of $M$. We have shown that $f$ is smooth. $\blacksquare$

Proof. Let $(\psi_M,A)$ and $(\psi_N,B)$ be charts of $M$ and $N$. Let $\hat f$ denote $\psi_N\circ f\circ\psi_M^{-1}:\psi_M(A\cap f^{-1}(B))\to\psi_N(B)$. Let $q\in\psi_M(A\cap f^{-1}(B))$ and let $p$ denote $\psi_M^{-1}(q)$, then $p\in A\cap f^{-1}(B)$. Thus there exist a smooth chart $(\varphi_M,U)$ where $p\in U$ and a smooth chart $(\varphi_N,V)$ where $f(p)\in V$ such that $f(U)\subseteq V$ and $\varphi_N\circ f\circ\varphi_M^{-1}:\varphi_M(U)\to\varphi_N(V)$ is smooth. Since $(\varphi_M,U)$ and $(\psi_M,A\cap f^{-1}(B))$ are compatible, $\varphi_M\circ\psi_M^{-1}$ is smooth. Thus $\varphi_N\circ f\circ\psi_M^{-1}:\psi_M(U\cap A\cap f^{-1}(B))\to\varphi_N(V\cap B)$ is smooth. Since $(\varphi_N,V)$ and $(\psi_N,B)$ are compatible, $\psi_N\circ\varphi_N^{-1}$ is smooth. Thus $\psi_N\circ f\circ\psi_M^{-1}:\psi_M(U\cap A\cap f^{-1}(B))\to\psi_N(V\cap B)$ is smooth. Since $p\in U\cap A\cap f^{-1}(B)$, $q=\psi_M(p)\in\psi_M(U\cap A\cap f^{-1}(B))$. Note that $\psi_M(U\cap A\cap f^{-1}(B))$ is open. We have shown that $\hat f$ is smooth. $\blacksquare$

Proof. Let $p\in M$, then $f(p)\in N$. Thus there exist a smooth chart $(\varphi_N,V)$ where $f(p)\in V$ and a smooth chart $(\varphi_P,W)$ where $g(f(p))\in W$ such that $g(V)\subseteq W$ and $\varphi_P\circ g\circ\varphi_N^{-1}:\varphi_N(V)\to\varphi_P(W)$ is smooth. Let $U$ denote $f^{-1}(V)$, then $U$ is open and $p\in U$. Let $(\varphi_M,S)$ be a smooth chart of $M$ where $p\in S$. Then $(\varphi_M,U\cap S)$ is a smooth chart of $M$ and $p\in U\cap S$. And $\varphi_N\circ f\circ\varphi_M^{-1}:\varphi_M(U\cap S)\to\varphi_N(V)$ is smooth. Thus $\varphi_P\circ(g\circ f)\circ\varphi_M^{-1}:\varphi_M(U\cap S)\to\varphi_P(W)$ is smooth. Note that $(g\circ f)(U\cap S)\subseteq W$. We have shown that $g\circ f:M\to P$ is smooth. $\blacksquare$

Proof. This follows directly from the above proposition. $\blacksquare$

Proof. $f|_U:U\to f(U)$ is trivially bijective. Since $f$ is a diffeomorphism, it is a homeomorphism. Thus $f(U)$ is an open subset of $N$, and thus defines an open submanifold of $N$. Let $p\in U$, then for some smooth chart $(\varphi,A)$ of $M$, $p\in A$, and for some smooth chart $(\psi,B)$ of $N$, $f(p)\in B$. Note that $(\psi,B\cap f(U))$ is a smooth chart of $f(U)$, as an open submanifold of $N$, and $f(p)\in B\cap f(U)$. Also note that $(\varphi,A\cap f^{-1}(B\cap f(U)))$ is a smooth chart of $U$, as an open submanifold of $M$, and $p\in A\cap f^{-1}(B\cap f(U))$. Since $f(A\cap f^{-1}(B\cap f(U)))\subseteq B\cap f(U)$, $\psi\circ f\circ\varphi^{-1}:\varphi(A\cap f^{-1}(B\cap f(U)))\to\psi(B\cap f(U))$ is a coordinate representation of $f$ and is thus smooth. We have shown that $f|_U$ is smooth. Note that $(f|_U)^{-1}=f^{-1}|_{f(U)}$, which is smooth for a symmetric argument. Hence $f|_U:U\to f(U)$ is a diffeomorphism. $\blacksquare$

Proof. Suppose for contradiction that $p\in\Int M$ and $f(p)\in\partial N$. Then there exist a smooth chart $(\varphi_M,U)$ of $M$ where $p\in U$ and $\varphi_M(U)$ is an open subset of $R^n$, and a smooth chart $(\varphi_N,V)$ of $N$ where $f(p)\in V$ and $\varphi_N(V)$ is an open subset of $H^n$, such that $f(U)=V$ and $\varphi_N(f(p))\in\partial H^n$. Since coordinate representations of smooth maps are smooth, $\varphi_N\circ f\circ\varphi_M^{-1}:\varphi_M(U)\to\varphi_N(V)$ is a diffeomorphism. Since $f(p)\in V$, $\varphi_N(f(p))\in\varphi_N(V)$. Thus $\varphi_N(V)\cap\partial H^n\neq\emptyset$. Since $\partial H^n\neq\emptyset$, $n\neq0$. Then by a lemma above, $\varphi_M(U)$ and $\varphi_N(V)$ are not diffeomorphic, a contradiction. Hence if $p\in\Int M$, then $f(p)\in\Int N$. By a symmetric argument, if $q\in\Int N$, then $f^{-1}(q)\in\Int M$.

If $y\in f(\Int M)$, then for some $x\in\Int M$, $f(x)=y$, thus $y\in\Int N$. If $y\in\Int N$, then $f^{-1}(y)\in\Int M$ and $f(f^{-1}(y))=y$, thus $y\in f(\Int M)$. Hence $f(\Int M)=\Int N$. If $y\in f(\partial M)$, then for some $x\in\partial M$, $f(x)=y$, then $y\notin\Int N$, thus $y\in\partial N$. If $y\in\partial N$, then $f^{-1}(y)\notin\Int M$, thus $f^{-1}(y)\in\partial M$ and $f(f^{-1}(y))=y$, implying $y\in f(\partial M)$. Hence $f(\partial M)=\partial N$.

Since $\Int M$ is an open submanifold of $M$, $f|_{\Int M}:\Int M\to\Int N$ is a diffeomorphism. $\blacksquare$

Proof. Suppose $M$ and $N$ are smooth manifolds without boundary. Since $M$ is non-empty, there exists $p\in M$. Since $M$ and $N$ are diffeomorphic, there exists a diffeomorphism $f:M\to N$. Then there exist a smooth chart $(\varphi_M,U)$ of $M$ where $p\in U$ and a smooth chart $(\varphi_N,V)$ of $N$ where $f(p)\in V$ such that $f(U)=V$. Since coordinate representations of smooth maps are smooth, $\varphi_N\circ f\circ\varphi_M^{-1}:\varphi_M(U)\to\varphi_N(V)$ is a diffeomorphism. Since $\varphi_M(U)$ is a non-empty open subset of $R^m$ and $\varphi_N(V)$ is a non-empty open subset of $R^n$, $m=n$.

Suppose $M$ and $N$ are smooth manifolds with boundary. Since $M$ and $N$ are diffeomorphic, there exists a diffeomorphism $f:M\to N$. Then $f|_{\Int M}:\Int M\to\Int N$ is a diffeomorphism. Thus $\Int M$ is a smooth $m$-manifold, $\Int N$ is a smooth $n$-manifold, and $\Int M$ and $\Int N$ are diffeomorphic. Since $M$ is non-empty, there exists $p\in M$. If $p\in\Int M$, then $\Int M$ is non-empty. If $p\in\partial M$, then for some boundary chart $(\varphi,U)$, $p\in U$ and $\varphi(p)\in\partial H^n$. Note that for some open subset $V$ of $R^n$, $V\cap H^n=\varphi(U)$. Since $\varphi(p)\in V$, for some $r\gt0$, $B_r(\varphi(p))\subseteq V$. Let $q$ denote $\varphi(p)+\frac{r}{2}\vb e_n$, then $B_{\frac{r}{2}}(q)\subseteq V\cap H^n=\varphi(U)$. Hence $(\varphi,\varphi^{-1}(B_{\frac{r}{2}}(q)))$ is an interior chart of $M$. Since $\varphi^{-1}(q)\in\varphi^{-1}(B_{\frac{r}{2}}(q))$, $\varphi^{-1}(q)\in\Int M$, implying $\Int M$ is non-empty. By a symmetric argument, since $N$ is non-empty, $\Int N$ is non-empty. Therefore, $m=n$. $\blacksquare$

Proof. Let $p\in\varphi^{-1}(\overline{B_s(0)(\cap H^n)})$. Then $\varphi(p)\in\overline{B_s(0)(\cap H^n)}$. Thus there exists a sequence $(q_i)$ of $B_s(0)(\cap H^n)$ that converges to $\varphi(p)$. And we can define a sequence $(p_i)$ of $\varphi^{-1}(B_s(0)(\cap H^n))$ such that $p_i=\varphi^{-1}(q_i)$. Let $U$ be a neighborhood of $p$, then it contains an open subset $V$ of $M$ such that $p\in V$. Note that $\varphi(V\cap B)$ is an open subset of $B_r(0)(\cap H^n)$ and $\varphi(p)\in\varphi(V\cap B)$, thus for some $t\gt0$, $B_t(\varphi(p))(\cap H^n)\subseteq\varphi(V\cap B)$. Since $(q_i)$ converges to $\varphi(p)$, for some $n\in N$, for all $k\ge n$, $q_k\in B_t(\varphi(p))(\cap H^n)$, implying $p_k=\varphi^{-1}(q_k)\in V\cap B\subseteq U$. We have shown that $(p_i)$ converges to $p$. Hence $p\in\overline{\varphi^{-1}(B_s(0)(\cap H^n))}$.

Let $p\in\overline{\varphi^{-1}(B_s(0)(\cap H^n))}$. Then there exists a sequence $(p_i)$ of $\varphi^{-1}(B_s(0)(\cap H^n))$ that converges to $p$. And we can define a sequence $(q_i)$ of $B_s(0)(\cap H^n)$ such that $q_i=\varphi(p_i)$. Since $(q_i)$ is bounded, it has a convergent subsequence $(q'_j)$. Let $q$ denote the limit of $(q'_j)$, then $q\in\overline{B_s(0)(\cap H^n)}$ and $\varphi^{-1}(q)\in B$. Then we can define a sequence $(p'_j)$ of $\varphi^{-1}(B_s(0)(\cap H^n))$ such that $p'_j=\varphi^{-1}(q'_j)$, which is a subsequence of $(p_i)$ and thus converges to $p$. By the same reasoning as the last paragraph, $(p'_j)$ converges to $\varphi^{-1}(q)$. Thus $p=\varphi^{-1}(q)\in\varphi^{-1}(\overline{B_s(0)(\cap H^n)})$. $\blacksquare$

Proof. Let $M$ be a smooth $n$-manifold with boundary. Then the domains of smooth charts of $M$ forms an open cover of $M$, which has countable subcover $\mathcal C$. Let $U\in\mathcal C$. Then there exists a smooth chart $(\varphi_U,U)$ of $M$. Let $\mathcal B_U$ denote the set of open balls $B_r(q)$ such that $r$ is rational, $q$ has rational coordinates, and $B_{2r}(q)\subseteq\varphi_U(U)$. And let $\mathcal B'_U$ denote

• $\emptyset$ if $(\varphi_U,U)$ is an interior chart;
• the set of open half balls $B_r(q)\cap H^n$ such that $r$ is rational, $q$ has rational coordinates, and $B_{2r}(q)\cap H^n\subseteq\varphi_U(U)$, if $(\varphi_U,U)$ is a boundary chart.

Define $\mathcal A_U$ to be $\{\varphi_U^{-1}(B):B\in\mathcal B_U\}$ and $\mathcal A'_U$ to be $\{\varphi_U^{-1}(B):B\in\mathcal B'_U\}$. Then $\bigcup_{U\in\mathcal C}\p{\mathcal A_U\cup\mathcal A'_U}$ is a countable set of subsets of $M$, which we will denote as $\mathcal A$.

Let $A\in\mathcal A$. Then for some $U\in\mathcal C$, either $A\in\mathcal A_U$ or $A\in\mathcal A'_U$.

• If $A\in\mathcal A_U$, then for some $B\in\mathcal B_U$, $A=\varphi_U^{-1}(B)$. Note that $B=B_r(q)$ for some rational $r$ and $q$ such that $B_{2r}(q)\subseteq\varphi_U(U)$. We define $A'$ to be $\varphi_U^{-1}(B_{2r}(q))$. Regardless of whether $(\varphi_U,U)$ is an interior chart or a boundary chart, $B_r(q)$ and $B_{2r}(q)$ are open subsets of $\varphi_U(U)$, thus $A$ and $A'$ are open subsets of $U$ and thus $M$. Define $h:R^n\to R^n$ such that $h(x)=x-q$, then $(h\circ\varphi_U,A')$ is a smooth chart, and we have $(h\circ\varphi_U)(A)=B_r(0)$ and $(h\circ\varphi_U)(A')=B_{2r}(0)$. Note that $$\overline A=\overline{(h\circ\varphi_U)^{-1}(B_r(0))}=(h\circ\varphi_U)^{-1}(\overline{B_r(0)})\subseteq(h\circ\varphi_U)^{-1}(B_{2r}(0))=A'$$ And we have $(h\circ\varphi_U)(\overline A)=\overline{B_r(0)}$. We have shown that $A$ is a regular ball.
• If $A\in\mathcal A'_U$, then for some $B\in\mathcal B'_U$, $A=\varphi_U^{-1}(B)$. Since $\mathcal B'_U$ is non-empty, $(\varphi_U,U)$ is a boundary chart, and $B=B_r(q)\cap H^n$ for some rational $r$ and $q$ such that $B_{2r}(q)\cap H^n\subseteq\varphi_U(U)$. We define $A'$ to be $\varphi_U^{-1}(B_{2r}(q)\cap H^n)$. Since $(\varphi_U,U)$ is a boundary chart, $B_r(q)\cap H^n$ and $B_{2r}(q)\cap H^n$ are open subsets of $\varphi_U(U)$, thus $A$ and $A'$ are open subsets of $U$ and thus $M$. Define $h:R^n\to R^n$ such that $h(x)=x-q$, then $(h\circ\varphi_U,A')$ is a smooth chart, and we have $(h\circ\varphi_U)(A)=B_r(0)\cap H^n$ and $(h\circ\varphi_U)(A')=B_{2r}(0)\cap H^n$. Note that $$\overline A=\overline{(h\circ\varphi_U)^{-1}(B_r(0)\cap H^n)}=(h\circ\varphi_U)^{-1}(\overline{B_r(0)\cap H^n})\subseteq(h\circ\varphi_U)^{-1}(B_{2r}(0)\cap H^n)=A'$$ And we have $(h\circ\varphi_U)(\overline A)=\overline{B_r(0)\cap H^n}$. We have shown that $A$ is a regular half ball.

Thus every element of $\mathcal A$ is a regular ball or a regular half ball.

Let $S$ be an open subset of $M$. Then $S=\cup\{S\cap U:U\in\mathcal C\}$. Let $U\in\mathcal C$, then $\varphi_U(S\cap U)$ is an open subset of $\varphi_U(U)$.

• Suppose $(\varphi_U,U)$ is an interior chart. Then $\varphi_U(S\cap U)$ is an open subset of $R^n$. Let $p\in\varphi_U(S\cap U)$, then for some $r\gt0$, $B_r(p)\subseteq\varphi_U(S\cap U)$. Note that there exists $q\in B_{\frac{r}{4}}(p)$ such that $q$ has rational coordinates, and there exists $s\in(\frac{r}{2},\frac{3r}{4})$ such that $s$ is rational. Then $B_{\frac{s}{2}}(q)$ is an open ball containing $p$ such that $\frac{s}{2}$ is rational, $q$ has rational coordinates, and $B_s(q)\subseteq B_r(p)\subseteq\varphi_U(S\cap U)\subseteq\varphi_U(U)$. Thus $p\in\cup\{B\in\mathcal B_U|B\subseteq\varphi_U(S\cap U)\}$. The other direction is trivial. Hence $\varphi_U(S\cap U)=\cup\{B\in\mathcal B_U|B\subseteq\varphi_U(S\cap U)\}$. Let $\mathcal B_{U,S}$ denote $\{B\in\mathcal B_U|B\subseteq\varphi_U(S\cap U)\}$, then $S\cap U=\cup\{\varphi_U^{-1}(B):B\in\mathcal B_{U,S}\}$. Note that $\{\varphi_U^{-1}(B):B\in\mathcal B_{U,S}\}\subseteq\mathcal A_U\subseteq\mathcal A_U\cup\mathcal A'_U\subseteq\mathcal A$. Thus $S\cap U$ is the union of a subset of $\mathcal A$.
• Suppose $(\varphi_U,U)$ is a boundary chart. Then $n\neq0$ and $\varphi_U(S\cap U)$ is an open subset of $H^n$. Let $p\in\varphi_U(S\cap U)$.
  • If $p\in\Int H^n$, then for some $r\gt0$, $B_r(p)\subseteq\varphi_U(S\cap U)$. By the same reasoning as above, $p\in\cup\{B\in\mathcal B_U|B\subseteq\varphi_U(S\cap U)\}$
  • If $p\in\partial H^n$, then for some $r\gt0$, $B_r(p)\cap H^n\subseteq\varphi_U(S\cap U)$. Note that there exists $q\in B_{\frac{r}{4}}(p)$ such that $q$ has rational coordinates and $q\in\partial H^n$, and there exists $s\in(\frac{r}{2},\frac{3r}{4})$ such that $s$ is rational. Then $B_{\frac{s}{2}}(q)\cap H^n$ is an open half ball containing $p$ such that $\frac{s}{2}$ is rational, $q$ has rational coordinates, and $B_s(q)\cap H^n\subseteq B_r(p)\cap H^n\subseteq\varphi_U(S\cap U)\subseteq\varphi_U(U)$. Thus $p\in\cup\{B\in\mathcal B'_U|B\subseteq\varphi_U(S\cap U)\}$.
  In either case, we have $p\in\cup\{B\in\mathcal B_U\cup\mathcal B'_U|B\subseteq\varphi_U(S\cap U)\}$. The other direction is trivial. Hence $\varphi_U(S\cap U)=\cup\{B\in\mathcal B_U\cup\mathcal B'_U|B\subseteq\varphi_U(S\cap U)\}$. Let $\mathcal B_{U,S}$ denote $\{B\in\mathcal B_U|B\subseteq\varphi_U(S\cap U)\}$ and $\mathcal B'_{U,S}$ denote $\{B\in\mathcal B'_U|B\subseteq\varphi_U(S\cap U)\}$, then $S\cap U=\cup\{\varphi_U^{-1}(B):B\in\mathcal B_{U,S}\cup\mathcal B'_{U,S}\}$. Note that $\{\varphi_U^{-1}(B):B\in\mathcal B_{U,S}\cup\mathcal B'_{U,S}\}\subseteq\mathcal A_U\cup\mathcal A'_U\subseteq\mathcal A$. Thus $S\cap U$ is the union of a subset of $\mathcal A$.

Since each $S\cap U$ is the union of a subset of $\mathcal A$, $S$ is the union of a subset of $\mathcal A$. Thus $\mathcal A$ is a basis of $M$.

Note that if $M$ is a smooth manifold without boundary, then every smooth chart is an interior chart, thus every element of $\mathcal A$ is a regular ball. $\blacksquare$

Proof.

• Let $p\in M$ then there exists an open set $B$ containing $p$ such that $\{S\in\mathcal C|S\cap B\neq\emptyset\}$, denoted $\mathcal D$, is finite. Let $\mathcal C'$ denote $\mathcal C\setminus\mathcal D$, then for all $S\in\mathcal C'$, we have $S\cap B=\emptyset$, thus $B\subseteq M\setminus S$, implying $B\subseteq\text{Ext} S$, and thus $\overline S\cap B=\emptyset$. Hence $\{S\in\mathcal C|\overline S\cap B\neq\emptyset\}\subseteq\mathcal D$, which is finite. Note that $S\mapsto\overline S$ defines a surjection from $\{S\in\mathcal C|\overline S\cap B\neq\emptyset\}$ to $\{S\in\{\overline U:U\in\mathcal C\}|S\cap B\neq\emptyset\}$. Therefore, $\{\overline U:U\in\mathcal C\}$ is locally finite.
  
  
• Let $p\in\overline{\cup\mathcal C}$. Then $p$ has a neighborhood $B_p$ such that $\{S\in\mathcal C|S\cap B_p\neq\emptyset\}$, denoted $\mathcal D$, is finite, and for every neighborhood $B$ of $p$, $\cup\mathcal C\cap(B\setminus\{p\})\neq\emptyset$. Suppose for contradiction that $p\notin\cup_{U\in\mathcal C}\overline U$, then for all $U\in\mathcal D$, $U\in\mathcal C$, then $p\notin\overline U$, thus $p\in\text{Ext} U$. Then either $\mathcal D$ is empty or $p\in\bigcap_{U\in\mathcal D}\text{Ext} U$, which is open. If $\mathcal D$ is empty, then $\cup\mathcal C\cap(B_p\setminus\{p\})=\emptyset$, a contradiction. Otherwise, $B_p\cap\bigcap_{U\in\mathcal D}\text{Ext} U$, denoted $B'_p$, is a neighborhood of $p$, and $\cup\mathcal C\cap(B'_p\setminus\{p\})=\emptyset$, a contradiction. Thus $p\in\cup_{U\in\mathcal C}\overline U$.
  
  Let $p\in\cup_{U\in\mathcal C}\overline U$. Then for some $U\in\mathcal C$, $p\in\overline U\subseteq\overline{\cup\mathcal C}$.

$\blacksquare$

Proof. Let $M$ be a smooth manifold with boundary, then it has a countable basis $\mathcal B$ of regular balls or regular half balls. Note that given an open ball $B$, $\overline B$ is closed and bounded, and thus compact in $R^n$. Similarly, given an open half ball $B$, $\overline B$ is compact in $R^n$, and thus compact in $H^n$. Let $p\in M$, then there exists $B\in\mathcal B$ such that $p\in B$, and $\overline B$ is a neighborhood of $p$. If $B$ is a regular ball, then there exists $B'\subseteq M$ such that $\overline B\subseteq B'$, and there exists a smooth chart $(\varphi,B')$ of $M$ such that for some $0\lt r\lt r'$, $\varphi(B)=B_r(0)$, $\varphi(\overline B)=\overline{B_r(0)}$, and $\varphi(B')=B_{r'}(0)$. Note that $\overline{B_r(0)}$ is compact in $\varphi(B')$. Since $\varphi^{-1}$ is continuous, $\overline B=\varphi^{-1}(\overline{B_r(0)})$ is compact in $B'$, and thus compact in $M$. If $B$ is a regular half ball, then there exists $B'\subseteq M$ such that $\overline B\subseteq B'$, and there exists a smooth chart $(\varphi,B')$ of $M$ such that for some $0\lt r\lt r'$, $\varphi(B)=B_r(0)\cap H^n$, $\varphi(\overline B)=\overline{B_r(0)\cap H^n}$, and $\varphi(B')=B_{r'}(0)\cap H^n$. Note that $\overline{B_r(0)\cap H^n}$ is compact in $\varphi(B')$. Since $\varphi^{-1}$ is continuous, $\overline B=\varphi^{-1}(\overline{B_r(0)\cap H^n})$ is compact in $B'$, and thus compact in $M$. We have shown that there exists a compact neighborhood of $p$. Hence $M$ is locally compact. $\blacksquare$

Proof. Let $X$ be such a space. Since $X$ is Hausdorff and locally compact, it has a precompact basis $\mathcal B$. Since $X$ is second-countable and $\mathcal B$ is an open cover of $X$, it has a countable subcover $\mathcal C$, which can be indexed into a sequence $(C_i)$ (extend with $\emptyset$ if $\mathcal C$ is finite) of open precompact sets that covers $X$. Suppose $A$ is a compact set and $n\in N$, then $(C_i)$ covers $A$ and thus has a finite subcover, implying for some $k\in N$ such that $k\ge n+1$, $A\subseteq\cup_{j=0}^kC_j\subseteq\cup_{j=0}^k\overline{C_j}$. Note that $\cup_{j=0}^k\overline{C_j}$ is compact, $n+1\in N$, and $A\subseteq\cup_{j=0}^kC_j\subseteq\Int\cup_{j=0}^k\overline{C_j}$. This defines a function $(A,n)\mapsto(\cup_{j=0}^k\overline{C_j},n+1)$ where $k\ge n+1$ from and onto pairs formed by a compact subsets of $X$ and a natural number. By recursion, we can define a sequence of pairs $(K_i,n_i)$ such that $(K_0,n_0)=(\overline{C_0},0)$ and $(K_{i+1},n_{i+1})=(\cup_{j=0}^k\overline{C_j},n_i+1)$ where $k\ge n_i+1$, and from which we can extract a sequence of compact sets $(K_i)$. Note that $n_i=i$ and $C_0\subseteq K_0$. Let $i\in N^+$, then $K_i=\cup_{j=0}^k\overline{C_j}$ where $k\ge n_{i-1}+1=i$, thus $C_i\subseteq K_i$. Hence $X\subseteq\cup_i C_i\subseteq\cup_i K_i\subseteq X$. Note that $K_i\subseteq\Int K_{i+1}$ by the definition of the recursion function. $\blacksquare$

Proof. Define $(K_i)$ as in the last lemma, then $X=\cup_i K_i$ and $K_i\subseteq\Int K_{i+1}$ for all $i\in N$. Let $K_{-1}=K_{-2}=\emptyset$ and define $V_j=K_j\setminus\Int K_{j-1}$ and $W_j=\Int K_{j+1}\setminus K_{j-2}$ for $j\in N$. Since $X$ is Hausdorff, compact sets are closed, thus each $V_j$ is closed and each $W_j$ is open. Then each $V_j$ is a closed subset of a compact set, and is thus compact. Note that $V_j\subseteq W_j$ for $j\in N$. Let $p\in X$, then there exists $j\in N$ such that $p\in K_j$. Thus there exists a least $k\in N$ such that $p\in K_k$, implying $p\in V_k$. Then we have $X=\cup_{j\in N}V_j$. Let $j\in N$, then for all $p\in V_j$, there exists $C_p\in\mathcal C$ such that $p\in C_p$. Note that $p\in C_p\cap W_j$, which is open, thus there exists $B_p\in\mathcal B$ such that $p\in B_p\subseteq C_p\cap W_j$. Then $\{B_p:p\in V_j\}$ is an open cover of $V_j$ and thus has a finite subcover $\mathcal B_j$, which is a subset of $\mathcal B$. Then $\cup_{j\in N}\mathcal B_j$, denoted $\mathcal B'$, is an open refinement of $\mathcal C$ that is a countable subset of $\mathcal B$. Note that for all $B\in\mathcal B'$, $B\subseteq W_j$ for some $j\in N$. Also note that $W_j\cap W_{j+k+3}=\emptyset$ for all $j,k\in N$. Let $p\in X$, then for some $B\in\mathcal B'$, $p\in B$, which is a neighborhood of $p$, and $B\subseteq W_j$ for some $j\in N$. Then $\{S\in\mathcal B'|B\cap S\neq\emptyset\}\subseteq\cup\{\mathcal B_k:k\in N,k\lt j+3,j\lt k+3\}$, which is finite. Hence $\mathcal B'$ is locally finite. $\blacksquare$

Proof. Note that by limit of composite functions, $$\lim_{t\to 0^+}e^{-\frac{1}{t}}=0$$ Let $k\in N$ and suppose $\lim_{t\to 0^+}e^{-\frac{1}{t}}t^{-k}=0$, then $$ \lim_{t\to 0^+}e^{-\frac{1}{t}}t^{-(k+1)} =\lim_{t\to 0^+}\frac{t^{-(k+1)}}{e^{\frac{1}{t}}} =\lim_{t\to 0^+}\frac{-(k+1)t^{-(k+2)}}{e^{\frac{1}{t}}(-t^{-2})} =\lim_{t\to 0^+}\frac{(k+1)t^{-k}}{e^{\frac{1}{t}}} =0 $$ By induction, for all $k\in N$, $\lim_{t\to 0^+}e^{-\frac{1}{t}}t^{-k}=0$.

Let $k\in N$ and suppose $$ f^{(k)}(t) = \begin{cases} e^{-\frac{1}{t}}\sum_{j=0}^{2k}c_jt^{-j} & t\gt0\\ 0 & t\le0 \end{cases} $$ for some $c_0,\ldots,c_{2k}\in R$. Note that for each $j$, with domain restricted to $R^+$, $$\p{t^{-j}e^{-\frac{1}{t}}}'=(-j)t^{-(j+1)}e^{-\frac{1}{t}}+t^{-(j+2)}e^{-\frac{1}{t}}$$ Thus $f^{(k+1)}$ takes the form $e^{-\frac{1}{t}}\sum_{j=0}^{2(k+1)}d_jt^{-j}$ for some $d_0,\ldots,d_{2(k+1)}\in R$ when $t\gt0$. Note that $$\lim_{h\to 0^+}\frac{f^{(k)}(h)-f^{(k)}(0)}{h}=\lim_{h\to 0^+}e^{-\frac{1}{h}}\sum_{j=0}^{2k}c_jh^{-(j+1)}=0$$ Thus $f^{(k+1)}(0)=0$, which concludes the inductive step. By induction, for all $k\in N$, $f^{(k)}$ exists on $R$. Hence $f$ is smooth. $\blacksquare$

Proof. Let $f$ be the function in the lemma above, then the function $$\frac{f(b-t)}{f(b-t)+f(t-a)}$$ clearly satisfies the conditions. $\blacksquare$

Proof. Let $H(x)=h(d(x,p))$, where $h$ is the function obtained in the lemma above with $a$ being $r_1$ and $b$ being $r_2$. Note that $d(x,p)=\norm{x-p}$ and the norm function is smooth on $R^n\setminus{0}$, thus $H(x)$ is smooth on $R^n\setminus\{p\}$. Since $H(x)$ is constant on $B_{r_1}(p)$, it is also smooth at $p$. $\blacksquare$

Proof. Let $a\in A$, then $U_a$ is an open submanifold of $M$, and thus has a basis $\mathcal B_a$ of regular balls or regular half balls. Let $\mathcal B$ denote $\cup\{\mathcal B_a:a\in A\}$, then $\mathcal B$ is a basis of $M$. Since $M$ is second-countable, locally compact, and Hausdorff, there exists a locally finite open refinement of $\{U_a:a\in A\}$ that is a countable subset of $\mathcal B$, which we will index into a bijection $(B_i)_{i\in k}$ where $k$ is countable.

Let $i\in k$, then for some $a\in A$, $B_i\in\mathcal B_a$, implying $B_i$ either a regular ball or a regular half ball of $U_a$. Then for some smooth ball $B'_i$ of $U_a$ where $\overline{B_i}\subseteq B'_i$, there exists a smooth chart $(\varphi_i,B'_i)$ of $U_a$ such that for some $0\lt r_i\lt r'_i$, $\varphi_i(B_i)=B_{r_i}(0)(\cap H^n)$, $\varphi_i(\overline{B_i})=\overline{B_{r_i}(0)(\cap H^n)}$, and $\varphi_i(B'_i)=B_{r'_i}(0)(\cap H^n)$. Let $H_i:R^n\to R$ be a smooth function such that $$ H_i(x) \begin{cases} =1 & d(x,0)\le\frac{r_i}{2}\\ \in(0,1) & d(x,0)\in(\frac{r_i}{2},r_i)\\ =0 & d(x,0)\ge r_i \end{cases} $$ Define $f_i:M\to R$ by $$ f_i(x)= \begin{cases} (H_i\circ\varphi_i)(x) & x\in B'_i\\ 0 & x\notin B'_i \end{cases} $$ Then $$ f_i(x) \begin{cases} \gt0 & x\in B_i\\ =0 & x\notin B_i \end{cases} $$ implying $\text{supp}f_i=\overline{B_i}$, and $f_i$ is clearly smooth.

Define $f:M\to R$ by $$f(p)=\sum_{i\in k}f_i(p)$$ Let $p\in M$, since $\{B_i:i\in k\}$ is locally finite and $B_i$ is injective, there exists an open subset $S$ of $M$ containing $p$ such that for some finite $j\subseteq k$, for all $i\in k$, $B_i\cap S\neq\emptyset$ if and only if $i\in j$. Thus for all $x\in S$, $f(x)=\sum_{i\in j}f_i(x)$. Note that for each $i\in j$, $f_i$ is smooth on the open submanifold $S$. Thus $f|_S$ is smooth. We have shown that $f$ is smooth. Let $p\in M$, then for some $i\in k$, $p\in B_i$, thus $f(p)\ge f_i(p)\gt 0$. Hence $f(M)\subseteq R^+$, implying $\frac{1}{f}$ is smooth, thus $$g_i=\frac{f_i}{f}$$ is smooth for each $i$. Clearly, let $p\in M$, then $\sum_{i\in k}g_i(p)=1$ and for each $i\in k$, $0\le g_i(p)\le 1$.

Let $i\in k$, then for some $a_i\in A$, $B'_i\subseteq U_{a_i}$. Now for each $a\in A$, let $j_a$ denote $\{i\in k|a_i=a\}$ and define $\psi_a:M\to R$ by $$\psi_a(p)=\sum_{i\in j_a}g_i(p)$$ Then $\psi_a$ is smooth and $0\le\psi_a(p)\le 1$ for all $p\in M$ and $a\in A$. Let $a\in A$, then $\{B_i:i\in j_a\}$ is locally finite, thus $$\text{supp}\psi_a=\overline{\bigcup_{i\in j_a}B_i}=\bigcup_{i\in j_a}\overline{B_i}\subseteq U_a$$ Let $p\in M$, we showed that there exists an open subset $S$ of $M$ containing $p$ such that for some finite $j\subseteq k$, for all $i\in k$, $B_i\cap S\neq\emptyset$ if and only if $i\in j$. If $i\in k\setminus j$, then $B_i\cap S=\emptyset$, thus $\overline{B_i}\cap S=\emptyset$. Hence for all $i\in k$, if $\overline{B_i}\cap S\neq\emptyset$, then $i\in j$. Let $j'$ denote $\{i\in j|\overline{B_i}\cap S\neq\emptyset\}$, then $a_i$ defines a surjection from $j'$ to $\{a\in A|\text{supp}\psi_a\cap S\neq\emptyset\}$, which is thus finite. We have shown that $\{\text{supp}\psi_a:a\in A\}$ is locally finite. Let $p\in M$, then $\{i\in k|g_i(p)\neq0\}$, denoted $l$, is finite. Let $A'$ denote $\{a_i:i\in l\}$, then $A'$ is finite, and $$\sum_{a\in A}\psi_a(p)=\sum_{a\in A}\sum_{i\in j_a}g_i(p)=\sum_{a\in A'}\sum_{i\in j_a\cap l}g_i(p)=\sum_{i\in l}g_i(p)=1$$ Therefore, $(\psi_a)_{a\in A}$ is a smooth partition of unity with respect to $(U_a)_{a\in A}$. $\blacksquare$

Proof. Define $S_0=\text{Ext}U$ and $S_1=V$, then $(S_a)_{a\in\{0,1\}}$ is an open cover of $M$. Thus there exists a smooth partition of unity $(\psi_a)_{a\in\{0,1\}}$ with respect to $(S_a)_{a\in\{0,1\}}$. Note that $\psi_1$ is a smooth real-valued function defined on $M$, $0\le\psi_1(p)\le 1$ for all $p\in M$, and $\psi_1$ is supported in $V$. Since $\psi_0$ is supported in $\text{Ext}U$, for all $p\in U$, $\psi_0(p)=0$, thus $\psi_1(p)=\sum_{a\in\{0,1\}}\psi_a(p)=1$. We have shown that $\psi_1$ is a smooth bump function for $U$ supported in $V$. $\blacksquare$

Proof. Let $p\in V$, then $p\in U$, thus there exists an open submanifold $U_p$ of $M$ containing $p$ and a smooth function $f_p:U_p\to N$ that agrees with $f$ on $U\cap U_p$. Let $x\in V\cap U_p$, then $f_p(x)=f(x)=f|_V(x)$. Thus $f_p$ agrees with $f|_V$ on $V\cap U_p$. Hence $f|_V$ is smooth. $\blacksquare$

Proof. Let $p\in U$, then there exists an open submanifold $V_p$ of $N$ containing $f(p)$ and a smooth function $g_p:V_p\to P$ that agrees with $g$ on $V\cap V_p$, and there exists an open submanifold $U_p$ of $M$ containing $p$ and a smooth function $f_p:U_p\to N$ that agrees with $f$ on $U\cap U_p$. Note that $f_p^{-1}(V_p)$ is open, thus $f_p|_{f_p^{-1}(V_p)}$ is smooth, and $f_p(p)=f(p)\in V_p$. Now we have an open submanifold $f_p^{-1}(V_p)$ of $M$ containing $p$ and a smooth function $g_p\circ f_p|_{f_p^{-1}(V_p)}$ such that for all $x\in U\cap f_p^{-1}(V_p)$, $(g_p\circ f_p|_{f_p^{-1}(V_p)})(x)=g_p(f_p(x))=g(f(x))=(g\circ f)(x)$. We have shown that $g\circ f$ is smooth. $\blacksquare$

Proof. Trivial. $\blacksquare$

Proof. Let $p\in U$, then there exists an open submanifold $U_p$ of $M$ such that $p\in U_p\subseteq V$ and a smooth function $f_p:U_p\to R^m$ that agrees with $f$ on $U\cap U_p$. Define $U'=U\cup\{U\}$ and $U_U=M\setminus U$, then $(U_p)_{p\in U'}$ is an open cover of $M$. Let $(\psi_p)_{p\in U'}$ be the corresponding smooth partition of unity. For each $p\in U$, $\text{supp}\psi_p\subseteq U_p$. Extend $f_p$ so that $f_p(x)=0$ for all $x\in M\setminus U_p$, then $\psi_pf_p:M\to R^m$ is smooth, and $\text{supp}\psi_pf_p\subseteq\text{supp}\psi_p$. Thus $\{\text{supp}\psi_pf_p:p\in U\}$ is locally finite. Hence we can define $F:M\to R^m$ by $$F(x)=\sum_{p\in U}(\psi_pf_p)(x)$$ Let $p\in M$, then there exists some open subset $B$ of $M$ containing $p$ and some finite $S\subseteq U$ such that $F|_B=\sum_{p\in S}(\psi_pf_p)|_B$, which is smooth. Thus $F$ is smooth. Note that $\text{supp}\psi_U\subseteq M\setminus U$, thus for all $x\in U$, $$F(x) =\sum_{\{p\in U|(\psi_pf_p)(x)\neq0\}}\psi_p(x)f_p(x) =\p{\sum_{\{p\in U|(\psi_pf_p)(x)\neq0\}}\psi_p(x)}f(x) =\p{\sum_{\{p\in U|\psi_p(x)\neq0\}}\psi_p(x)}f(x) =\p{\sum_{\{p\in U'|\psi_p(x)\neq0\}}\psi_p(x)}f(x) =f(x) $$ Let $x\in M$ such that $F(x)\neq0$. Then there exists $p\in U$ such that $(\psi_pf_p)(x)\neq0$, implying $\psi_p(x)\neq0$, and thus $x\in\text{supp}\psi_p\subseteq\bigcup_{p\in U}\text{supp}\psi_p$. Note that $\{\text{supp}\psi_p:p\in U\}$ is locally finite. Hence $$\text{supp}F\subseteq\overline{\bigcup_{p\in U}\text{supp}\psi_p}=\bigcup_{p\in U}\overline{\text{supp}\psi_p}=\bigcup_{p\in U}\text{supp}\psi_p\subseteq V$$ We have shown that $F$ is the desired function. $\blacksquare$

Proof. Suppose $p$ is an interior (boundary) point, then there exists an interior (a boundary) chart $(\varphi,S)$ such that $p\in S\subseteq U$. Since $\varphi(S)$ is open, we can find an open (half) ball $B$ centered at $\varphi(p)$ such that $\overline B\subseteq\varphi(S)$. Note that $\overline B$ is a compact subset of $\varphi(S)$. Then $\varphi^{-1}(\overline B)$ is a compact subset of $S$ and thus $M$, implying it is closed. Note that $p\in\varphi^{-1}(B)\subseteq\varphi^{-1}(\overline B)\subseteq S\subseteq U$. Hence $\varphi^{-1}(\overline B)$ is a closed neighborhood of $p$ contained in $U$. $\blacksquare$

Proof. Since $U$ is a neighborhood of $p$, there exists an open subset $B$ of $M$ such that $p\in B\subseteq U$. By the lemma above, there exists a closed neighborhood $U^*$ of $p$ contained in $B$, and thus contained in $U$. Let $f:U\to R^m$ be smooth, then $f|_{U^*}$ is smooth. Thus there exists a smooth function $\overline f:M\to R^m$ that agrees with $f|_{U^*}$ on $U^*$, and hence agrees with $f$ on $U^*$. $\blacksquare$

Proof. Let $B$ be a regular ball containing $p$. Since $\varphi(B'\setminus\{p\})=\varphi(B')\setminus\{\varphi(p)\}$ is open, $B'\setminus\{p\}$ is open. Thus $M\setminus\{p\}=(B'\setminus\{p\})\cup(M\setminus\overline B)$ is open. $\blacksquare$

Proof. Let $f_1\in C^\infty(M)$ be a constant function mapping to $1$. Then $v(f_1)=v(f_1f_1)=f_1(p)v(f_1)+f_1(p)v(f_1)=2v(f_1)$, thus $v(f_1)=0$. Let $c$ be the constant that $f$ maps to, then $v(f)=v(cf_1)=cv(f_1)=0$.

If $f(p)=g(p)=0$, then $v(fg)=f(p)v(g)+g(p)v(f)=0+0=0$.

Suppose $f$ agrees with $g$ in a neighborhood $N$ of $p$. Let $S$ be an open subset of $M$ such that $p\in S\subseteq N$. Note that $M\setminus\{p\}$ is open and $\overline{M\setminus S}=M\setminus S\subseteq M\setminus\{p\}$. Thus there exists a smooth bump function $h$ for $M\setminus S$ supported in $M\setminus\{p\}$. Note that $(f-g)h=f-g$ and $h(p)=(f-g)(p)=0$. Thus $v(f-g)=v((f-g)h)=0$, implying $v(f)=v(g)$. $\blacksquare$

Proof. Suppose $v\in R^n$. Let $f,g\in C^\infty(R^n)$ and $c\in R$, then $$D_v|_p(cf)=\eval{\dv{}{t}}_{t=0}(cf)(p+tv)=D(cf)(p)D(p+tv)(0)=cDf(p)D(p+tv)(0)=c\eval{\dv{}{t}}_{t=0}f(p+tv)=cD_v|_p(f)$$ $$D_v|_p(f+g)=\eval{\dv{}{t}}_{t=0}(f+g)(p+tv)=D(f+g)(p)D(p+tv)(0)=(Df(p)+Dg(p))D(p+tv)(0)=\eval{\dv{}{t}}_{t=0}f(p+tv)+\eval{\dv{}{t}}_{t=0}g(p+tv)=D_v|_p(f)+D_v|_p(g)$$ $$D_v|_p(fg)=\eval{\dv{}{t}}_{t=0}(fg)(p+tv)=D(fg)(p)D(p+tv)(0)=(f(p)Dg(p)+g(p)Df(p))D(p+tv)(0)=f(p)\eval{\dv{}{t}}_{t=0}(g)(p+tv)+g(p)\eval{\dv{}{t}}_{t=0}(f)(p+tv)=f(p)D_v|_p(g)+g(p)D_v|_p(f)$$ Thus $D_v|_p\in T_pR^n$.

Suppose $w\in T_pR^n$. Define $v$ to be $(w(x_1),\ldots,w(x_n))$. Let $f\in C^\infty(R^n)$. By generalized Taylor's theorem with $k=1$, for all $x\in R^n$, $$f(x)=f(p)+\sum_iD_if(p)(x_i-p_i)+\sum_{i,j}(x_i-p_i)(x_j-p_j)\int_0^1D_jD_if(p+t(x-p))(1-t^k)dt$$ Note that given $i,j\in\{1,\ldots,n\}$, $$(x_i-p_i)|_{x=p}=\left.\p{(x_j-p_j)\int_0^1D_jD_if(p+t(x-p))(1-t^k)dt}\right|_{x=p}=0$$ Thus the first and last term are mapped to $0$ by $w$. And we have $$w(f)=w\p{\sum_iD_if(p)(x_i-p_i)}=\sum_iD_if(p)w(x_i-p_i)=\sum_iD_if(p)v_i=D_v|_p(f)$$ Thus $w=D_v|_p$.

We have shown that $T_pR^n=\{D_v|_p:v\in R^n\}$.

Let $v,w\in R^n$ and $c\in R$. Let $f\in C^\infty(R^n)$, then $$D_{cv}|_p(f)=\eval{\dv{}{t}}_{t=0}f(p+t(cv))=D(f)(p)D(p+t(cv))(0)=D(f)(p)(cD(p+tv))(0)=c\eval{\dv{}{t}}_{t=0}f(p+tv)=cD_v|_p(f)$$ $$D_{v+w}|_p(f)=\eval{\dv{}{t}}_{t=0}f(p+t(v+w))=D(f)(p)D(p+t(v+w))(0)=D(f)(p)(D(p+tv)+D(p+tw))(0)=\eval{\dv{}{t}}_{t=0}f(p+tv)+\eval{\dv{}{t}}_{t=0}f(p+tw)=D_v|_p(f)+D_w|_p(f)$$ Thus let $w\in T_pR^n$, there exists $v\in R^n$ such that $$w=D_v|_p=D_{\sum_iv_i\vb e_i}|_p=\sum_iv_iD_{\vb e_i}|_p$$ Hence the span of $(D_{\vb e_1}|_p,\ldots,D_{\vb e_n}|_p)$ is $T_pR^n$.

Let $c_1,\ldots,c_n\in R$ such that $\sum_ic_iD_{\vb e_i}|_p=0$. Then for each $j\in\{1,\ldots,n\}$, $$c_j=\sum_ic_iD_{\vb e_i}|_p(x_j)=\p{\sum_ic_iD_{\vb e_i}|_p}(x_j)=0$$ Thus $(D_{\vb e_1}|_p,\ldots,D_{\vb e_n}|_p)$ is independent.

We have shown that $(D_{\vb e_1}|_p,\ldots,D_{\vb e_n}|_p)$ is a basis of $T_pR^n$. Thus $T_pR^n$ is an $n$-dimensional vector space. $\blacksquare$

Proof. Let $f,g\in C^\infty(N)$, then $w_v(f+g)=v((f+g)\circ F)=v(f\circ F+g\circ F)=v(f\circ F)+v(g\circ F)=w_v(f)+w_v(g)$.

Let $f\in C^\infty(N)$ and $c\in R$, then $w_v(cf)=v((cf)\circ F)=v(c(f\circ F))=cv(f\circ F)=cw_v(f)$.

Let $f,g\in C^\infty(N)$, then $w_v(fg)=v(fg\circ F)=v((f\circ F)(g\circ F))=(f\circ F)(p)v(g\circ F)+(g\circ F)(p)v(f\circ F) =f(F(p))w_v(g)+g(F(p))w_v(f)$. $\blacksquare$

Proof.

• • Let $v,w\in T_pM$, let $f\in C^\infty(N)$, then $dF_p(v+w)(f)=(v+w)(f\circ F)=v(f\circ F)+w(f\circ F)=dF_p(v)(f)+dF_p(w)(f)=(dF_p(v)+dF_p(w))(f)$. Thus $dF_p(v+w)=dF_p(v)+dF_p(w)$.
  • Let $v\in T_pM$, let $c\in R$, let $f\in C^\infty(N)$, then $dF_p(cv)(f)=(cv)(f\circ F)=c(v(f\circ F))=c(dF_p(v)(f))=(cdF_p(v))(f)$. Thus $dF_p(cv)=cdF_p(v)$.
  We have shown that $dF_p$ is linear.
• Let $v\in T_pM$, let $f\in C^\infty(P)$, then $d(G\circ F)_p(v)(f)=v(f\circ G\circ F)=dF_p(v)(f\circ G)=dG_{F(p)}(dF_p(v))(f)=(dG_{F(p)}\circ dF_p)(v)(f)$. Thus $d(G\circ F)_p(v)=(dG_{F(p)}\circ dF_p)(v)$. Hence $d(G\circ F)_p=dG_{F(p)}\circ dF_p$.
• Suppose $F$ is a diffeomorphism. Then both $F$ and $F^{-1}$ are smooth. Thus both $dF_p$ and $dF^{-1}_{F(p)}$ are linear, implying they are both homomorphisms. Note that $dF^{-1}_{F(p)}\circ dF_p=d(F^{-1}\circ F)_p=d(I_M)_p=I_{T_pM}$ and $dF_p\circ dF^{-1}_{F(p)}=d(F\circ F^{-1})_{F(p)}=d(I_N)_{F(p)}=I_{T_{F(p)}N}$. Thus $dF_p$ is bijective and $(dF_p)^{-1}=dF^{-1}_{F(p)}$. We have shown that $dF_p$ is an isomorphism.

$\blacksquare$

Proof. Let $h=f-g$, then $h(x)=0$ in an open set $U$ containing $p$ and $h(p)=0$. Note that $\overline{\{p\}}=\{p\}\subseteq U$, and thus $\overline{M\setminus U}=M\setminus U\subseteq M\setminus\{p\}=M\setminus\overline{\{p\}}$, which is open. Thus there exists a smooth bump function $\varphi$ for $M\setminus U$ supported in $M\setminus\{p\}$. Since $p\in M\setminus\{\text{supp}\varphi\}$, $\varphi(p)=0$. Note that $h,\varphi\in C^\infty(M)$, thus $v(h\varphi)=0$. Also note that $h=h\varphi$, thus $v(f)-v(g)=v(h)=0$, and hence $v(f)=v(g)$. $\blacksquare$

Proof. Note that there exists a neighborhood $U^*\subseteq U$ of $p$, such that for all $f\in C^\infty(U)$, there exists $\overline f\in C^\infty(M)$ that agrees with $f$ on $U^*$.

Suppose $v,w\in T_pU$ and $dI_p(v)=dI_p(w)$. Let $f\in C^\infty(U)$, then $v(f)=v(\overline f|_U)=v(\overline f\circ I)=dI_p(v)(\overline f)$. Similarly, we have $w(f)=dI_p(w)(\overline f)$. Thus $v(f)=w(f)$. Hence $v=w$. We have shown that $dI_p$ is injective.

Let $w\in T_pM$. Then $f\mapsto w(\overline f)$ defines a function $v$ from $C^\infty(U)$ to $R$. Note that for all $f,g\in C^\infty(U)$ and $c\in R$, $w(\overline{f+g})=w(\overline f+\overline g)$, $w(\overline{cf})=w(c\overline f)$, and $w(\overline{fg})=w(\overline f\overline g)$. Thus $v$ is clearly linear, and for all $f,g\in C^\infty(U)$, $v(fg)=w(\overline f\overline g)=\overline f(p)w(\overline g)+\overline g(p)w(\overline f)=f(p)v(g)+g(p)v(f)$. Hence $v\in T_pU$. Note that for all $f\in C^\infty(M)$, $dI_p(v)(f)=v(f\circ I)=w(\overline{f\circ I})=w(f)$. Thus $dI_p(v)=w$. We have shown that $dI_p$ is surjective.

Since $dI_p$ is linear and bijective, it is an isomorphism. $\blacksquare$

Proof. Since we assumed $x\in\partial H^n$, we have $n\neq0$. Since $H^n$ is a closed subset of $R^n$, for all $f\in C^\infty(H^n)$, there exists $\overline f\in C^\infty(R^n)$ that agrees with $f$ on $H^n$, and $f=\overline f\circ I$.

Suppose $v,w\in T_xH^n$ and $dI_x(v)=dI_x(w)$. Let $f\in C^\infty(H^n)$, then $v(f)=v(\overline f\circ I)=dI_x(v)(\overline f)=dI_x(w)(\overline f)=w(\overline f\circ I)=w(f)$. Thus $v=w$. We have shown that $dI_x$ is injective.

Let $w\in T_xR^n$. Then for some $c_1,\ldots,c_n\in R$, $w=\sum_ic_iD_{\vb e_i}|_x$. Given $f,g\in C^\infty(R^n)$ such that $f$ and $g$ agree on $H^n$, for all $i\in\{1,\ldots,n-1\}$, $f(x+t\vb e_i)=g(x+t\vb e_i)$ for $t\in R$. And $f(x+t\vb e_n)=g(x+t\vb e_n)$ for $t\ge0$, implying their one-sided derivatives at $0$ approaching from the right are equal, and thus their derivatives at $0$ are equal. Thus $$w(f)=\sum_ic_iD_{\vb e_i}|_x(f)=\sum_ic_i\eval{\dv{}{t}}_{t=0}f(x+t\vb e_i) =\sum_ic_i\eval{\dv{}{t}}_{t=0}g(x+t\vb e_i)=\sum_ic_iD_{\vb e_i}|_x(g)=w(g)$$ Hence for all $f,g\in C^\infty(H^n)$ and $c\in R$, $w(\overline{f+g})=w(\overline f+\overline g)$, $w(\overline{cf})=w(c\overline f)$, and $w(\overline{fg})=w(\overline f\overline g)$. Note that $f\mapsto w(\overline f)$ defines a function $v$ from $C^\infty(H^n)$ to $R$. $v$ is clearly linear, and for all $f,g\in C^\infty(H^n)$, $v(fg)=w(\overline f\overline g)=\overline f(x)w(\overline g)+\overline g(x)w(\overline f)=f(x)v(g)+g(x)v(f)$. Hence $v\in T_xH^n$. Note that for all $f\in C^\infty(R^n)$, $dI_x(v)(f)=v(f\circ I)=w(\overline{f\circ I})=w(f)$. Thus $dI_x(v)=w$. We have shown that $dI_x$ is surjective.

Since $dI_x$ is linear and bijective, it is an isomorphism. $\blacksquare$

Proof. If $p$ is an interior chart, then there exists an interior chart $(\varphi,U)$ where $p\in U$. Note that $\varphi:U\to\varphi(U)$ is a diffeomorphism, thus $d\varphi_p$ is an isomorphism, hence $T_pM\cong T_pU\cong T_{\varphi(p)}\varphi(U)\cong T_{\varphi(p)}R^n$. Since $T_{\varphi(p)}R^n$ is $n$-dimensional, $T_pM$ is $n$-dimensional.

If $p$ is a boundary chart, then there exists a boundary chart $(\varphi,U)$ where $p\in U$ and $\varphi(p)\in\partial H^n$. Note that $\varphi:U\to\varphi(U)$ is a diffeomorphism, thus $d\varphi_p$ is an isomorphism, hence $T_pM\cong T_pU\cong T_{\varphi(p)}\varphi(U)\cong T_{\varphi(p)}H^n\cong T_{\varphi(p)}R^n$. Since $T_{\varphi(p)}R^n$ is $n$-dimensional, $T_pM$ is $n$-dimensional. $\blacksquare$

Proof. Suppose $\varphi$ is an interior chart. We showed that $d(I_U)_p\circ d(\varphi^{-1})_{\hat p}\circ(d(I_{\hat U})_{\hat p})^{-1}:T_{\hat p}R^n\to T_pM$ is an isomorphism and $(\pdv{}{x_i}|_{\hat p})_i$ is a basis of $T_{\hat p}R^n$. Thus $((d(I_U)_p\circ d(\varphi^{-1})_{\hat p}\circ(d(I_{\hat U})_{\hat p})^{-1})(\pdv{}{x_i}|_{\hat p}))_i$ is a basis of $T_pM$. We showed that given $w\in T_{\hat p}R^n$, $f\mapsto w(\overline f)$ defines a function $v_w\in T_{\hat p}\hat U$, where $\overline f$ agrees with $f$ on a neighborhood ${\hat U}^*\subseteq\hat U$ of $\hat p$. Then $w\mapsto v_w$ defines a function from $T_{\hat p}R^n$ to $T_{\hat p}\hat U$, which happens to be $(d(I_{\hat U})_{\hat p})^{-1}$. Thus given $w\in T_{\hat p}R^n$ and $f\in C^\infty(\hat U)$, $(d(I_{\hat U})_{\hat p})^{-1}(w)(f)=w(\overline f)$. Note that $(d(I_{\hat U})_{\hat p})^{-1}(\pdv{}{x_i}|_{\hat p})(f)=\pdv{\overline f}{x_i}(\hat p)=\pdv{f}{x_i}(\hat p)$. Thus $(d(I_{\hat U})_{\hat p})^{-1}(\pdv{}{x_i}|_{\hat p})$ will simply be denoted $\pdv{}{x_i}|_{\hat p}$. Given $f\in C^\infty(U)$, $d(\varphi^{-1})_{\hat p}(\pdv{}{x_i}|_{\hat p})(f)=\pdv{\hat f}{x_i}(\hat p)$, And we denote $d(\varphi^{-1})_{\hat p}(\pdv{}{x_i}|_{\hat p})$ as $\pdv{}{x_i}|_p$. Given $f\in C^\infty(M)$, $d(I_U)_p(\pdv{}{x_i}|_p)(f)=\pdv{}{x_i}|_p(f|_U)$. Thus we will denote $d(I_U)_p(\pdv{}{x_i}|_p)$ simply as $\pdv{}{x_i}|_p$. And $(\pdv{}{x_i}|_p)_i$ is a basis of $T_pM$.

Suppose $\varphi$ is a boundary chart. We showed that $d(I_U)_p\circ d(\varphi^{-1})_{\hat p}\circ(d(I_{\hat U})_{\hat p})^{-1}\circ(d(I_{H^n})_{\hat p})^{-1}:T_{\hat p}R^n\to T_pM$ is an isomorphism and $(\pdv{}{x_i}|_{\hat p})_i$ is a basis of $T_{\hat p}R^n$. Thus $((d(I_U)_p\circ d(\varphi^{-1})_{\hat p}\circ(d(I_{\hat U})_{\hat p})^{-1}\circ(d(I_{H^n})_{\hat p})^{-1})(\pdv{}{x_i}|_{\hat p}))_i$ is a basis of $T_pM$. We showed that given $w\in T_{\hat p}R^n$, $f\mapsto w(\overline f)$ defines a function $v_w\in T_{\hat p}H^n$, where $\overline f$ agrees with $f$ on $H^n$. Then $w\mapsto v_w$ defines a function from $T_{\hat p}R^n$ to $T_{\hat p}H^n$, which happens to be $(d(I_{H^n})_{\hat p})^{-1}$. Thus given $w\in T_{\hat p}R^n$ and $f\in C^\infty(H^n)$, $(d(I_{H^n})_{\hat p})^{-1}(w)(f)=w(\overline f)$. Note that $(d(I_{H^n})_{\hat p})^{-1}(\pdv{}{x_i}|_{\hat p})(f)=\pdv{\overline f}{x_i}(\hat p)=\pdv{f}{x_i}(\hat p)$, where when $p\in\partial H^n$ and $i=n$, the partial derivative is one-sided approaching from the right. The rest of the proof for the basis is similar to above.

Now suppose $v\in T_pM$. Then for some real $(v_i)_i$, $v=v_i\eval{\pdv{}{x_i}}_p$. Thus for each $j$, $$v(\overline{x_j\circ\varphi}) =v_i\pdv{((\overline{x_j\circ\varphi})\circ\varphi^{-1})}{x_i}(\varphi(p)) =v_i\pdv{x_j}{x_i}(\varphi(p)) =v_j $$ where $\overline{x_j\circ\varphi}$ is a local extension of $x_j\circ\varphi$. $\blacksquare$

Proof. Let $S$ be an open set of $M$ containing $p$ such that $(\varphi,U)$ and $(\psi,V)$ agree on $S$. Then $\varphi(S)$ and $\psi(S)$ are open and equivalent. Let $B$ be an open ball (or open half ball if $p$ is a boundary point) centered at $\varphi(p)$ and contained by $\varphi(S)$. Then for all $x\in B$, $\varphi^{-1}(x)=\psi^{-1}(x)$, thus for each $i$ and for all $f\in C^\infty(M)$, $\eval{\pdv{}{x_i}}_p(f)=\pdv{f\circ\varphi^{-1}}{x_i}(\varphi(p))=\pdv{f\circ\psi^{-1}}{y_i}(\psi(p))=\eval{\pdv{}{y_i}}_p(f)$. Hence $\eval{\pdv{}{x_i}}_p=\eval{\pdv{}{y_i}}_p$ for each $i$. $\blacksquare$

Proof. Let $f\in C^\infty(M)$, then $$ dI_p\p{\eval{\pdv{}{x_i}}_p^U}(f) =\eval{\pdv{}{x_i}}_p^U(f|_U) =\pdv{\hat{f|_U}}{x_i}(\hat p) =\pdv{\hat{f}}{x_i}(\hat p) =\eval{\pdv{}{x_i}}_p^M(f) $$ $\blacksquare$

Proof. When either $M$ or $N$ is $0$-dimensional, this is trivial. Now suppose otherwise.

Let $S=U\cap F^{-1}(V)$, then $\varphi|_S:S\to\varphi(S)$ is a smooth chart of $M$ containing $p$. For each $i$, $$ d(F|_S)_p\p{\eval{\pdv{}{x_i}}_p^S} =(d(F|_S)_p\circ d(\varphi|_S^{-1})_{\hat p})\p{\eval{\pdv{}{x_i}}_{\hat p}} =d(F|_S\circ\varphi|_S^{-1})_{\hat p}\p{\eval{\pdv{}{x_i}}_{\hat p}} =d(\psi^{-1}\circ\hat F)_{\hat p}\p{\eval{\pdv{}{x_i}}_{\hat p}} =(d\psi^{-1}_{\hat F(\hat p)}\circ d\hat F_{\hat p})\p{\eval{\pdv{}{x_i}}_{\hat p}} $$ Let $f\in C^\infty(\psi(V))$. In case $p$ or $F(p)$ is a boundary point, we define smooth $\overline{\hat F}$ containing $\hat p$ that agrees with $\hat F$ on the intersection of their domains and smooth $\overline f$ containing $\hat F(\hat p)$ that agrees with $f$ on the intersection of their domains, such that $\overline f\circ\overline{\hat F}$ agrees with $f\circ\hat F$ on the intersection of their domains. Then $$ d\hat F_{\hat p}\p{\eval{\pdv{}{x_i}}_{\hat p}}(f) =\pdv{(f\circ\hat F)}{x_i}(\hat p) =\pdv{(\overline f\circ\overline{\hat F})}{x_i}(\hat p) =\pdv{\overline f}{y_j}(\hat F(\hat p))\pdv{\overline{\hat F}_j}{x_i}(\hat p) =\pdv{f}{y_j}(\hat F(\hat p))\pdv{\hat F_j}{x_i}(\hat p) =\p{\pdv{\hat F_j}{x_i}(\hat p)\eval{\pdv{}{y_j}}_{\hat F(\hat p)}}(f) $$ Thus $$ d(F|_S)_p\p{\eval{\pdv{}{x_i}}_p^S} =d\psi^{-1}_{\hat F(\hat p)}\p{\pdv{\hat F_j}{x_i}(\hat p)\eval{\pdv{}{y_j}}_{\hat F(\hat p)}} =\pdv{\hat F_j}{x_i}(\hat p)\eval{\pdv{}{y_j}}_{F(p)}^V $$ Then given $f\in C^\infty(N)$, $$ dF_p\p{\eval{\pdv{}{x_i}}_p}(f) =\eval{\pdv{}{x_i}}_p(f\circ F) =\eval{\pdv{}{x_i}}_p^S(f|_V\circ F|_S) =d(F|_S)_p\p{\eval{\pdv{}{x_i}}_p^S}(f|_V) =\pdv{\hat F_j}{x_i}(\hat p)\eval{\pdv{}{y_j}}_{F(p)}^V(f|_V) =\pdv{\hat F_j}{x_i}(\hat p)\eval{\pdv{}{y_j}}_{F(p)}(f) $$ Note that $\eval{\pdv{}{x_i}}_p$ is defined with $\varphi|_S$ instead of $\varphi$. However, $\eval{\pdv{}{x_i}}_p$ defined with $\varphi|_S$ is equivalent to that defined with $\varphi$. Hence $$dF_p\p{\eval{\pdv{}{x_i}}_p}=\pdv{\hat F_j}{x_i}(\hat p)\eval{\pdv{}{y_j}}_{F(p)}$$ with respect to $\varphi$ and $\psi$. And entry $(j,i)$ of the matrix representation of $dF_p$ is $$\pdv{\hat F_j}{x_i}(\hat p)$$ $\blacksquare$

Proof. Let $S$ denote $U\cap V$. Then for each $i$, $$ \eval{\pdv{}{x_i}}_p^S =d(\varphi|_S^{-1})_{\varphi(p)}\p{\eval{\pdv{}{x_i}}_{\varphi(p)}} =(d(\psi|_S^{-1})_{\psi(p)}\circ d(\psi|_S\circ\varphi|_S^{-1})_{\varphi(p)})\p{\eval{\pdv{}{x_i}}_{\varphi(p)}} =d(\psi|_S^{-1})_{\psi(p)}\p{\pdv{(\psi|_S\circ\varphi|_S^{-1})_j}{x_i}(\varphi(p))\eval{\pdv{}{y_j}}_{\psi(p)}} =\pdv{(\psi\circ\varphi^{-1})_j}{x_i}(\varphi(p))\eval{\pdv{}{y_j}}_p^S $$ Let $v\in T_pM$ with components $(v_i)$ with respect to $\varphi$ and $(v'_j)$ with respect to $\psi$. Since standard bases defined with $\varphi|_S,\psi|_S$ are equivalent to those defined with $\varphi,\psi$ respectively, we have $$ v'_j\eval{\pdv{}{y_j}}_p =v_i\eval{\pdv{}{x_i}}_p =v_idI_p\p{\eval{\pdv{}{x_i}}_p^S} =v_idI_p\p{\pdv{(\psi\circ\varphi^{-1})_j}{x_i}(\varphi(p))\eval{\pdv{}{y_j}}_p^S} =v_i\pdv{(\psi\circ\varphi^{-1})_j}{x_i}(\varphi(p))dI_p\p{\eval{\pdv{}{y_j}}_p^S} =v_i\pdv{(\psi\circ\varphi^{-1})_j}{x_i}(\varphi(p))\eval{\pdv{}{y_j}}_p $$ Thus for each $j$, $$v'_j=\pdv{(\psi\circ\varphi^{-1})_j}{x_i}(\varphi(p))v_i$$ $\blacksquare$

Proof. Let $v\in dI_p(T_p\partial M)$, then for some $w\in T_p\partial M$, $dI_p(w)=v$. Then $$v_n=v(\overline{x_n\circ\varphi})=w(\overline{x_n\circ\varphi}|_{\partial M})=w(0)=0$$ Now suppose $v_n=0$. Note that $dI_p\p{\eval{\pdv{}{x_i}}_p^{\partial M}}=\eval{\pdv{}{x_i}}_p^M$ for all $i\in\{1,\ldots,n-1\}$. Thus $$dI_p\p{v_i\eval{\pdv{}{x_i}}_p^{\partial M}}=v_i\eval{\pdv{}{x_i}}_p^M=v$$ implying $v\in dI_p(T_p\partial M)$.

Let $v$ be an inward-pointing. Then $v_n\neq0$ and there exists a smooth curve $\gamma:[0,\varepsilon)\to M$ such that $\gamma(0)=p$ and $\gamma'(0)=v$ for some $\varepsilon\gt0$. Note that $$ v =d\gamma_0\p{\eval{\dv{}{t}}_0} =\dv{\hat\gamma_i}{t}(0)\eval{\pdv{}{x_i}}_p $$ Thus $v_n=\dv{\hat\gamma_n}{t}(0)$, which is a one-sided derivative. Note that $\hat\gamma_n$ is always non-negative and $\hat\gamma_n(0)=0$, implying $v_n\ge0$. Since $v_n$ is also non-zero, we have $v_n\gt0$. Now suppose $v_n\gt0$. Then $v\in T_pM\setminus dI_p(T_p\partial M)$ and for some small enough $\varepsilon$, we can define $\gamma:[0,\varepsilon)\to M$ such that $\gamma(t)=\varphi^{-1}(\varphi(p)+(tv_1,\ldots,tv_n))$, which is a smooth curve. Then $\gamma(0)=p$ and $$\gamma'(0)=\dv{\hat\gamma_i}{t}(0)\eval{\pdv{}{x_i}}_p=v_i\eval{\pdv{}{x_i}}_p=v$$ Thus $v$ is inward-pointing.

Let $v$ be an outward-pointing. Then $v_n\neq0$ and there exists a smooth curve $\gamma:(-\varepsilon,0]\to M$ such that $\gamma(0)=p$ and $\gamma'(0)=v$ for some $\varepsilon\gt0$. Note that $$ v =-d\gamma_0\p{\eval{\dv{}{t}}_0} =-\dv{\hat\gamma_i}{t}(0)\eval{\pdv{}{x_i}}_p $$ Thus $v_n=-\dv{\hat\gamma_n}{t}(0)$, which is a one-sided derivative. Note that $\hat\gamma_n$ is always non-negative and $\hat\gamma_n(0)=0$, implying $v_n\le0$. Since $v_n$ is also non-zero, we have $v_n\lt0$. Now suppose $v_n\lt0$. Then $v\in T_pM\setminus dI_p(T_p\partial M)$ and for some small enough $\varepsilon$, we can define $\gamma:(-\varepsilon,0]\to M$ such that $\gamma(t)=\varphi^{-1}(\varphi(p)+(tv_1,\ldots,tv_n))$, which is a smooth curve. Then $\gamma(0)=p$ and $$\gamma'(0)=-\dv{\hat\gamma_i}{t}(0)\eval{\pdv{}{x_i}}_p=v_i\eval{\pdv{}{x_i}}_p=v$$ Thus $v$ is outward-pointing. $\blacksquare$

Proof. If $n=0$, this is trivial. Suppose below that $n\neq0$. Let $(\varphi,S)$ be a smooth chart of $M$ centered at $p$ and $(\psi,T)$ be a smooth chart of $N$ centered at $F(p)$ such that $F(S)\subseteq T$. Note that $\varphi:S\to\varphi(S)$ and $\psi:T\to\psi(T)$ are diffeomorphisms. Thus $d\hat F_0=d\psi_{F(p)}\circ dF_p\circ d\varphi^{-1}_0$ is bijective. Note that the matrix representation of $d\hat F_0$ is the total derivative $D\hat F(0)$, which is invertible. By inverse function theorem, there exists an open subset $W$ of $\hat S$ such that $0\in W$, $\hat F(W)$ is open, and $\hat F|_W:W\to\hat F(W)$ is a diffeomorphism. Restrict $W$ to an open ball $B$ centered at $0$, then $\hat F(B)$ is open and $\hat F|_B:B\to\hat F(B)$ is still a diffeomorphism. Note that $\varphi^{-1}(B)$ has all the desired properties of $U$. $\blacksquare$

Proof. Let $(\varphi,U)$ be a smooth chart of $M$. Clearly, $\varphi^*$ is bijective, and $\varphi^*(T^k_lTM|_U)=R^{n^{(k+l)}}\times\varphi(U)$ is an open subset of $R^{n^{(k+l)}+n}$ or $H^{n^{(k+l)}+n}$. Now let $(\varphi,U)$ and $(\psi,V)$ be smooth charts of $M$. Then $(\varphi,U\cap V)$ and $(\psi,U\cap V)$ are smooth charts of $M$, and $\varphi^{-1}\circ\psi$ is diffeomorphic. Clearly, $\psi^*\circ{\varphi^*}^{-1}:R^{n^{(k+l)}}\times\varphi(U\cap V)\to R^{n^{(k+l)}}\times\psi(U\cap V)$ is diffeomorphic, and thus homeomorphic.

Let $\tau$ denote the collection of subsets $S$ of $T^k_lTM$, such that $S$ is the union of a subset of $\cup_U\tau_U$. Trivially, $\emptyset\in\tau$. Note that $T^k_lTM|_U\in\tau_U$ for each smooth chart $(\varphi,U)$, thus $T^k_lTM\in\tau$. Let $\mathcal S$ be a subset of $\tau$. Then for all $S\in\mathcal S$, $S=\cup V_S$ where $V_S$ is a subset of $\cup_U\tau_U$. Then $\cup_SV_S$ is a subset of $\cup_U\tau_U$, and $\cup\mathcal S=\cup\cup_SV_S$, thus $\cup\mathcal S\in\tau$.

Let $(\varphi,U)$ and $(\psi,V)$ be smooth charts of $M$. Let $S,T$ be open subsets of $R^{n^{(k+l)}}\times\varphi(U)$ and $R^{n^{(k+l)}}\times\psi(V)$, and thus open subsets of $R^{n^{(k+l)}+n}$ or $H^{n^{(k+l)}+n}$. Note that $R^{n^{(k+l)}}\times\psi(U\cap V)$ is an open subset of $R^{n^{(k+l)}+n}$ or $H^{n^{(k+l)}+n}$, thus $T\cap(R^{n^{(k+l)}}\times\psi(U\cap V))$ is an open subset of $R^{n^{(k+l)}+n}$ or $H^{n^{(k+l)}+n}$, and thus an open subset of $R^{n^{(k+l)}}\times\psi(U\cap V)$. Since $\psi^*\circ{\varphi^*}^{-1}$ is homeomorphic, $(\varphi^*\circ{\psi^*}^{-1})(T\cap(R^{n^{(k+l)}}\times\psi(U\cap V)))$ is an open subset of $R^{n^{(k+l)}}\times\varphi(U\cap V)$, and thus an open subset of $R^{n^{(k+l)}}\times\varphi(U)$. Hence $S\cap(\varphi^*\circ{\psi^*}^{-1})(T\cap(R^{n^{(k+l)}}\times\psi(U\cap V)))$ is an open subset of $R^{n^{(k+l)}}\times\varphi(U)$. Thus ${\varphi^*}^{-1}(S\cap(\varphi^*\circ{\psi^*}^{-1})(T\cap(R^{n^{(k+l)}}\times\psi(U\cap V))))$ is an element of $\tau_U$, and hence an element of $\tau$. Note that $$ {\varphi^*}^{-1}(S\cap(\varphi^*\circ{\psi^*}^{-1})(T\cap(R^{n^{(k+l)}}\times\psi(U\cap V)))) ={\varphi^*}^{-1}(S)\bigcap{\varphi^*}^{-1}((\varphi^*\circ{\psi^*}^{-1})(T\cap(R^{n^{(k+l)}}\times\psi(U\cap V)))) $$ $$ ={\varphi^*}^{-1}(S)\bigcap{\psi^*}^{-1}(T\cap(R^{n^{(k+l)}}\times\psi(U\cap V))) ={\varphi^*}^{-1}(S)\bigcap{\psi^*}^{-1}(T)\bigcap{\psi^*}^{-1}((R^{n^{(k+l)}}\times\psi(U\cap V))) ={\varphi^*}^{-1}(S)\bigcap{\psi^*}^{-1}(T) $$
Let $\alpha,\beta\in\tau$, then $\alpha=\cup\eta_\alpha$ and $\beta=\cup\eta_\beta$ where $\eta_\alpha,\eta_\beta$ are subsets of $\cup_U\tau_U$. Let $S\in\eta_\alpha$ and $T\in\eta_\beta$, then for some smooth charts $(\varphi,U)$ and $(\psi,V)$ of $M$, $S\in\tau_U$ and $T\in\tau_V$. Then for some open subset $S^*$ of $R^{n^{(k+l)}}\times\varphi(U)$, $S={\varphi^*}^{-1}(S^*)$, and for some open subset $T^*$ of $R^{n^{(k+l)}}\times\psi(V)$, $T={\psi^*}^{-1}(T^*)$. Then $S\cap T={\varphi^*}^{-1}(S^*)\cap{\psi^*}^{-1}(T^*)\in\tau$. Hence $$ \alpha\cap\beta =\p{\bigcup\eta_\alpha}\bigcap\p{\bigcup\eta_\beta} =\bigcup\{S\cap T:(S,T)\in\eta_\alpha\times\eta_\beta\} \in\tau $$ We have shown that $\tau$ is a topology of $T^k_lTM$.

Let $(\varphi,U)$ be a smooth chart of $M$. Given an open subset $V^*$ of $R^{n^{(k+l)}}\times\varphi(U)$, ${\varphi^*}^{-1}(V^*)$ is an open subset of $T^k_lTM$, and thus an open subset of $T^k_lTM_U$. Hence $\varphi^*$ is continuous. Given an open subset $U^*$ of $T^k_lTM_U$, then $U^*\in\tau$, thus $U^*$ is the union of a subset $\mathcal S$ of $\cup_V\tau_V$. Let $S\in\mathcal S$, then for some smooth chart $(\psi,V)$ of $M$, $S\in\tau_V$, implying $S\subseteq T^k_lTM_{U\cap V}$ and $\psi^*(S)$ is an open subset of $R^{n^{(k+l)}}\times\psi(V)$. Since $\psi^*\circ{\varphi^*}^{-1}$ is homeomorphic, $\varphi^*(S)$ is an open subset of $R^{n^{(k+l)}}\times\varphi(U)$. Thus $\varphi^*(U^*)=\cup_S\varphi^*(S)$ is an open subset of $R^{n^{(k+l)}}\times\varphi(U)$. Hence ${\varphi^*}^{-1}$ is continuous. We have shown that $\varphi^*$ is a homeomorphism.

Let $p,q\in T^k_lTM$ be distinct. If $p_0=q_0$, then $p_1\neq q_1$, implying for some coordinate $i$, ${p_1}_i\neq{q_1}_i$. Let $A$ and $B$ be open subsets of $R^{n^{(k+l)}}$ split at $({p_1}_i+{q_1}_i)/2$ on coordinate $i$. Let $(U,\varphi)$ be a smooth chart such that $p_0=q_0\in U$. Then ${\varphi^*}^{-1}(A\times\varphi(U))$ and ${\varphi^*}^{-1}(B\times\varphi(U))$ are disjoint open subsets of $T^k_lTM$, one containing $p$ and the other containing $q$. If $p_0\neq q_0$, then there exist open subsets $S_p,S_q$ of $M$ such that $p_0\in S_p$, $q_0\in S_q$, and $S_p$ and $S_q$ are disjoint. Note that there exists smooth charts $(\varphi_p,U_p)$ and $(\varphi_q,U_q)$ of $M$ such that $p_0\in U_p$ and $q_0\in U_q$. Then $T^k_lTM|_{S_p\cap U_p}$ and $T^k_lTM|_{S_q\cap U_q}$ are disjoint open subsets of $T^k_lTM$ such that $p\in T^k_lTM|_{S_p\cap U_p}$ and $q\in T^k_lTM|_{S_q\cap U_q}$. We have shown that $T^k_lTM$ is Hausdorff.

Since $M$ is second-countable, there exists a sequence of smooth charts $(\varphi_i,U_i)$ that covers $M$. Note that $R^{n^{(k+l)}}\times\varphi_i(U_i)$ is second countable for each $i$, and we will denote the countable basis of the topology of $R^{n^{(k+l)}}\times\varphi_i(U_i)$ by $(\tau_{i,j})$. Note that $({\varphi_i^*}^{-1}(\tau_{i,j}))$ is a countable basis of the topology of $T^k_lTM$. We have shown that $T^k_lTM$ is second-countable.

Let $p\in T^k_lTM$. Then $p_0\in M$, and thus $p_0\in U$ for some smooth chart $(\varphi,U)$ of $M$. Hence $p\in T^k_lTM_U$. And $T^k_lTM_U$ is homeomorphic to $R^{n^{(k+l)}}\times\varphi(U)$ by $\varphi^*$. We have shown that $T^k_lTM$ is locally half-Euclidean.

We have shown that $T^k_lTM$ is a manifold with boundary. Trivially, the collection of maps $(\varphi^*,T^k_lTM|_U)$ for each smooth chart $(\varphi,U)$ of $M$ is an atlas of $T^k_lTM$, which generates a smooth structure on $T^k_lTM$. It is also trivial that the interior of $T^k_lTM$ is $T^k_lTM|_{\Int M}$ and the boundary of $T^k_lTM$ is $T^k_lTM|_{\partial M}$. $\blacksquare$

Proof. We will assume below that there exists a smooth chart $(\varphi,U)$ of $M$. The general case follows immediately.

Let $(e_k)$ denote the standard frame on $U$ and $(\mu^l)$ the standard coframe. Then $F_i=a_i^ke_k$ and $G^j=b^j_l\mu^l$ where each $a_i^k$ and $b^j_l$ is a real-valued function on $U$. Note that for all $p\in U$, $(a_i^k(p))_{ki}$ is the transition matrix from $({F_1}_p,\ldots,{F_n}_p)$ to $({e_1}_p,\ldots,{e_n}_p)$, and $(b^j_l(p))_{jl}$ is the transition matrix from $({G^1}_p,\ldots,{G^n}_p)$ to $({\mu^1}_p,\ldots,{\mu^n}_p)$. Thus $(a_i^k(p))_{ki}(b^j_l(p))_{jl}=I$. Hence every $a_i^k\circ\varphi^{-1}$ is smooth if and only if every $b^j_l\circ\varphi^{-1}$ is smooth. Thus every $F_i$ is smooth if and only if every $G^j$ is smooth. $\blacksquare$

Proof. Suppose $\omega$ is smooth. Let $(\varphi,U)$ be a smooth chart of $M$. Then the coordinate representation $\varphi^*\circ\omega\circ\varphi^{-1}$ is smooth. Let $\omega_U$ denote $\omega$ transformed into a tensor field on $U$. Then the coordinate representation of $\omega_U$ is the same as that of $\omega$ and is thus smooth. Thus the coordinate representation of each $\omega_I$ is trivially smooth. Since the coordinate representation is global, each $\omega_I$ is smooth.

Now suppose for each smooth chart $(\varphi,U)$ of $M$ and each $I$, $\omega_I$ is smooth. Given a smooth chart $(\varphi,U)$, for each $I$, the coordinate representation of $\omega_I$ is smooth. Trivially, the coordinate representation of $\omega_U$ is smooth. Since this coordinate representation of $\omega_U$ is the same as that of $\omega$, the latter is also smooth. This is sufficient to conclude that $\omega$ is smooth. $\blacksquare$

Proof. Let $(\varphi,U)$ be a smooth chart of $M$. Then $\omega_I\circ\varphi^{-1},{\omega_1}_I\circ\varphi^{-1},{\omega_2}_I\circ\varphi^{-1}$ are smooth for each $I$, $\eta_J\circ\varphi^{-1}$ is smooth for each $J$, and $c\circ\varphi^{-1}$ is smooth.

Note that $\omega_1+\omega_2=({\omega_1}_I+{\omega_2}_I)e^I$ with respect to $(\varphi,U)$. Since each $({\omega_1}_I+{\omega_2}_I)\circ\varphi^{-1}={\omega_1}_I\circ\varphi^{-1}+{\omega_2}_I\circ\varphi^{-1}$ is smooth, the coordinate representation of $\omega_1+\omega_2$ on $U$ is smooth.

Note that $c\omega=c\omega_Ie^I$ with respect to $(\varphi,U)$. Since each $(c\omega_I)\circ\varphi^{-1}=(c\circ\varphi^{-1})(\omega_I\circ\varphi^{-1})$ is smooth, the coordinate representation of $c\omega$ on $U$ is smooth.

We have shown that $\omega_1+\omega_2,c\omega\in\Omega^k(M)$. Now define the same notations.

Note that $\omega\wedge\eta=\p{\omega_Ie^I}\bigwedge\p{\eta_Je^J}=\omega_I\eta_J e^{IJ}$ with respect to $(\varphi,U)$. Since each $(\omega_I\eta_J)\circ\varphi^{-1}=(\omega_I\circ\varphi^{-1})(\eta_J\circ\varphi^{-1})$ is smooth, $\omega_I\eta_J\in C^\infty(U)$. Since each $e^{IJ}$ is clearly a $(k+l)$-form on $U$, $\omega_I\eta_J e^{IJ}$ is a $(k+l)$-form on $U$. Thus the coordinate representation of $\omega\wedge\eta$ on $U$ is smooth.

We have shown that $\omega\wedge\eta\in\Omega^{k+l}(M)$. $\blacksquare$

Proof. Let $p\in M$. Let $J$ be any $k$-tuple of $n$. If $J$ is not injective, then both $\omega^U_p(e^U_{J_1},\ldots,e^U_{J_k})$ and $\omega_p(dI_p(e^U_{J_1}),\ldots,dI_p(e^U_{J_k}))$ are $0$. If $J$ is injective, then $J=I\sigma$ for some strictly increasing $k$-tuple $I$ of $n$ and $\sigma\in S_k$. Then $$ \omega^U_p(e^U_{J_1},\ldots,e^U_{J_k}) =\text{sgn}(\sigma)\omega^U_p(e^U_{I_1},\ldots,e^U_{I_k}) =\text{sgn}(\sigma)\omega_p(dI_p(e^U_{I_1}),\ldots,dI_p(e^U_{I_k})) =\omega_p(dI_p(e^U_{J_1}),\ldots,dI_p(e^U_{J_k})) $$ Thus we have $\omega^U_p(e^U_{J_1},\ldots,e^U_{J_k})=\omega_p(dI_p(e^U_{J_1}),\ldots,dI_p(e^U_{J_k}))$. Let $v_1,\ldots,v_k\in T_pU$, then $$ \omega^U_p(v_1,\ldots,v_k) =\omega^U_p({v_1}^j{e^U}_j,\ldots,{v_k}^j{e^U}_j) =\omega_p({v_1}^jdI_p({e^U}_j),\ldots,{v_k}^jdI_p({e^U}_j)) =\omega_p(dI_p(v_1),\ldots,dI_p(v_k)) =(I^*\omega)_p(v_1,\ldots,v_k) $$ Thus $\omega^U=I^*\omega$. $\blacksquare$

Proof. $$ F^*(\omega_1+\omega_2)_p(v_1,\ldots,v_k) =(\omega_1+\omega_2)_{F(p)}(dF_p(v_1),\ldots,dF_p(v_k)) ={\omega_1}_{F(p)}(dF_p(v_1),\ldots,dF_p(v_k))+{\omega_2}_{F(p)}(dF_p(v_1),\ldots,dF_p(v_k)) =(F^*\omega_1)_p(v_1,\ldots,v_k)+(F^*\omega_2)_p(v_1,\ldots,v_k) =(F^*\omega_1+F^*\omega_2)_p(v_1,\ldots,v_k) $$ $$ F^*(c\omega)_p(v_1,\ldots,v_k) =(c\omega)_{F(p)}(dF_p(v_1),\ldots,dF_p(v_k)) =c(F(p))\omega_{F(p)}(dF_p(v_1),\ldots,dF_p(v_k)) =(c\circ F)(p)(F^*\omega)_p(v_1,\ldots,v_k) =((c\circ F)F^*\omega)_p(v_1,\ldots,v_k) $$ $$ F^*(\omega\wedge\eta)_p(v_1,\ldots,v_{k+l}) =(\omega\wedge\eta)_{F(p)}(dF_p(v_1),\ldots,dF_p(v_{k+l})) =\frac{1}{k!l!}\sum_{\sigma\in S_{k+l}}\text{sgn}(\sigma)(\omega_{F(p)}\otimes\eta_{F(p)})(dF_p(v_{\sigma_1}),\ldots,dF_p(v_{\sigma_{k+l}})) =\frac{1}{k!l!}\sum_{\sigma\in S_{k+l}}\text{sgn}(\sigma)((F^*\omega)_p\otimes(F^*\eta)_p)(v_{\sigma_1},\ldots,v_{\sigma_{k+l}}) =(F^*\omega\wedge F^*\eta)_p(v_1,\ldots,v_{k+l}) $$ $\blacksquare$

Proof. Let $p^*\in M$, then there exists a smooth chart $(\varphi,U)$ of $M$ and a smooth chart $(\psi,V)$ of $N$ such that $p^*\in U$, $F(p^*)\in V$, $F(U)\subseteq V$, and $\psi\circ F\circ\varphi^{-1}$ is smooth.

Let $\nu^j$ denote the tensor field of the dual basis $j$ on $V$. Then $$ F^*\nu^j_p(v) =\nu^j_{F(p)}(dF_p(v)) =dF_p(v)^j =dF_p(v)\p{\overline{y^j\circ\psi}} =v\p{\overline{y^j\circ\psi}\circ F} =v^i\eval{\pdv{}{x^i}}_p\p{\overline{y^j\circ\psi}\circ F} =\pdv{\p{\overline{y^j\circ\psi}\circ F\circ\varphi^{-1}}}{x^i}(\varphi(p))\mu^i_p(v) =\pdv{\p{y^j\circ\psi\circ F\circ\varphi^{-1}}}{x^i}(\varphi(p))\mu^i_p(v) $$ Let $c^j_i$ denote the real-valued map on $U$ defined by $\pdv{\p{y^j\circ\psi\circ F\circ\varphi^{-1}}}{x^i}(\varphi(p))$, then $c^j_i\in C^\infty(U)$, and we have $$F^*\nu^j=c^j_i\mu^i$$ implying $F^*\nu^j\in\Omega^1(U)$.

Since $$ F^*\omega =F^*\p{\omega_J\p{\bigwedge\nu^{J_j}}} =(\omega_J\circ F)\p{\bigwedge F^*\nu^{J_j}} $$ $F^*\omega\in\Omega^k(U)$. Note that given $v\in T_pU$, transforming $v$ by $T_pU\to T_{F(p)}V\to T_{F(p)}N$ or by $T_pU\to T_pM\to T_{F(p)}N$ results in the same element of $T_{F(p)}N$. Thus $F^*\omega$ understood as pullback of a tensor field on $V$ is the same as $F^*\omega$ understood as pullback of a tensor field on $N$ transformed into a tensor field on $U$. Hence the coordinate representation of $F^*\omega$, as pullback of a tensor field on $N$, on $U$ is smooth. This is sufficient to conclude that $F^*\omega\in\Omega^k(M)$. $\blacksquare$

Proof. $$ (F^*G^*\omega)_p(v_1,\ldots,v_k) =(G^*\omega)_{F(p)}(dF_p(v_1),\ldots,dF_p(v_k)) =\omega_{(G\circ F)(p)}((dG_{F(p)}\circ dF_p)(v_1),\ldots,(dG_{F(p)}\circ dF_p)(v_k)) =\omega_{(G\circ F)(p)}(d(G\circ F)_p(v_1),\ldots,d(G\circ F)_p(v_k)) =(G\circ F)^*\omega_p(v_1,\ldots,v_k) $$ $\blacksquare$

Proof. Let $p\in U$. Then $$ {(F^*\omega)_p}_I =(F^*\omega)_p(e_p^{I_1},\ldots,e_p^{I_k}) =\omega_{F(p)}(dF_p(e_p^{I_1}),\ldots,dF_p(e_p^{I_k})) =\omega_{F(p)}({dF_p}_{(j,{I_1})}e_{F(p)}^j,\ldots,{dF_p}_{(j,{I_k})}e_{F(p)}^j) =\prod_l{dF_p}_{({J_{\sigma_l}},{I_l})}\omega_{F(p)}(e_{F(p)}^{J_{\sigma_1}},\ldots,e_{F(p)}^{J_{\sigma_k}}) =\text{sgn}(\sigma)\prod_l({dF_p}_{JI})_{(\sigma_l,l)}\omega_{F(p)}(e_{F(p)}^{J_1},\ldots,e_{F(p)}^{J_k}) =\det({dF_p}^J_I){\omega_{F(p)}}_J $$ Where $J$ sums over strictly increasing $k$-tuples of $n$.

When transformed to $U$ and $V$, $$(F^*\omega)_I(p)=\det({dF_p}^J_I){\omega_J}(F(p))$$ Thus $$F^*\omega=(F^*\omega)_Ie^I=\det(DF)^J_I(\omega_J\circ F)e^I$$ $\blacksquare$

Proof. $$ d(\omega_1+\omega_2) =\pdv{(\omega_1+\omega_2)_I}{x^j}dx^j\wedge dx^I =\pdv{({\omega_1}_I+{\omega_2}_I)}{x^j}dx^j\wedge dx^I =\pdv{{\omega_1}_I}{x^j}dx^j\wedge dx^I+\pdv{{\omega_2}_I}{x^j}dx^j\wedge dx^I =d\omega_1+d\omega_2 $$ $$ d(c\omega) =\pdv{(c\omega)_I}{x^j}dx^j\wedge dx^I =\pdv{(c{\omega}_I)}{x^j}dx^j\wedge dx^I =c\pdv{({\omega}_I)}{x^j}dx^j\wedge dx^I =cd\omega $$ $$ d(\omega\wedge\eta) =d\p{\p{\omega_Idx^I}\wedge\p{\eta_Jdx^J}} =d\p{\omega_I\eta_Jdx^I\wedge dx^J} =d(\omega_I\eta_J)\wedge dx^I\wedge dx^J =\pdv{\omega_I\eta_J}{x^k}dx^k\wedge dx^I\wedge dx^J $$ $$ =\p{\pdv{\omega_I}{x^k}\eta_J+\pdv{\eta_J}{x^k}\omega_I}dx^k\wedge dx^I\wedge dx^J =\p{\pdv{\omega_I}{x^k}dx^k\wedge dx^I}\wedge\p{\eta_Jdx^J}+(-1)^k\p{\omega_Idx^I}\wedge\p{\pdv{\eta_J}{x^k}dx^k\wedge dx^J} =d\omega\wedge\eta+(-1)^k\omega\wedge d\eta $$ $$ d(d\omega) =d\p{D_j\omega_Idx^j\wedge dx^I} =d\p{D_j\omega_I}\wedge dx^j\wedge dx^I =D_kD_j\omega_Idx^k\wedge dx^j\wedge dx^I =\sum_{j\lt k}\p{D_kD_j\omega_Idx^k \wedge dx^j\wedge dx^I+D_jD_k\omega_Idx^j \wedge dx^k\wedge dx^I} =\sum_{j\lt k}\p{D_kD_j\omega_I-D_jD_k\omega_I}dx^k \wedge dx^j\wedge dx^I =0 $$ $\blacksquare$

Proof. $$ (F^*df)_p(v) =df_{F(p)}(dF_p(v)) =\pdv{f}{y^j}(F(p))dy^j_{F(p)}(dF_p(v)) =\pdv{f}{y^j}(F(p))dF_p(v)^j =\pdv{f}{y^j}(F(p))dF_p(v)(\overline{y^j}) =\pdv{f}{y^j}(F(p))v(\overline{y^j}\circ F) =\pdv{f}{y^j}(F(p))v^i\pdv{(\overline{y^j}\circ F)}{x^i}(p) =\pdv{f}{y^j}(F(p))\pdv{F^j}{x^i}(p)dx^i_p(v) =\pdv{(f\circ F)}{x^i}(p)dx^i_p(v) =d(f\circ F)_p(v) $$ $\blacksquare$

Proof. $$ F^*d\omega =F^*\p{d\omega_I\wedge\p{\bigwedge dy^{I_i}}} =d(\omega_I\circ F)\wedge\p{\bigwedge d(y^{I_i}\circ F)} =d\p{(\omega_I\circ F)\p{\bigwedge d(y^{I_i}\circ F)}} =dF^*\p{\omega_I\p{\bigwedge dy^{I_i}}} =dF^*\omega $$ $\blacksquare$

Proof. Let $(\varphi,U),(\psi,U)$ be smooth charts of $M$. Then $$ \varphi^*d{\varphi^{-1}}^*\omega =\psi^*{\psi^{-1}}^*\varphi^*d{\varphi^{-1}}^*\omega =\psi^*(\varphi\circ\psi^{-1})^*d{\varphi^{-1}}^*\omega =\psi^*d(\varphi\circ\psi^{-1})^*{\varphi^{-1}}^*\omega =\psi^*d{\psi^{-1}}^*\varphi^*{\varphi^{-1}}^*\omega =\psi^*d{\psi^{-1}}^*\omega $$ Now let $V$ be an open subset of $U$. Let $\omega_U$ denote $\omega$ transformed into a tensor field on $U$ and let $\omega_V$ denote $\omega$ transformed into a tensor field on $V$, then $\omega_U$ and $\omega_V$ share the same coordinates in $V$. Let $\varphi_U$ denote $\varphi$ and $\varphi_V$ denote $\varphi|_V:V\to\varphi(V)$. Since ${\omega_U}_I$ agrees with ${\omega_V}_I$ on $V$ for each $I$, $({\varphi_U^{-1}}^*\omega_U)_I$ agrees with $({\varphi_V^{-1}}^*\omega_V)_I$ on $\varphi(V)$ for each $I$. Thus their partial derivatives agree on $\varphi(V)$, and $(d{\varphi_U^{-1}}^*\omega_U)_J$ agrees with $(d{\varphi_V^{-1}}^*\omega_V)_J$ on $\varphi(V)$ for each $J$. Again, $({\varphi_U}^*d{\varphi_U^{-1}}^*\omega_U)_J$ agrees with $({\varphi_V}^*d{\varphi_V^{-1}}^*\omega_V)_J$ on $V$ for each $J$. Thus for any $p\in V$, $({\varphi_U}^*d{\varphi_U^{-1}}^*\omega_U)_p$ and $({\varphi_V}^*d{\varphi_V^{-1}}^*\omega_V)_p$ share the same coordinates, and when transformed back into tensors of $T_pM$, they are the same tensor. Combining what we have, we have shown that, given smooth charts $(\varphi,U),(\psi,V)$ of $M$, $$\p{\varphi^*d{\varphi^{-1}}^*\omega}_p=\p{\psi^*d{\psi^{-1}}^*\omega}_p$$ for all $p\in U\cap V$.

Let $(\varphi,U)$ be a smooth chart of $M$. Note that for all $p\in U$, $$ \p{\varphi^*d{\varphi^{-1}}^*\omega}_p =\p{\varphi_p^*d{\varphi_p^{-1}}^*\omega}_p =(d\omega)_p $$ Thus the coordinate representation of $d\omega$ is the same as that of $\varphi^*d{\varphi^{-1}}^*\omega$, which is smooth. Hence $d\omega\in\Omega^{k+1}(M)$. $\blacksquare$

Proof. Let $\omega\in\Omega^k(M)$. Let $d'\omega$ denote exterior derivative of $\omega$ on Euclidean space. Then for all $p\in M$, $$(d\omega)_p=\p{I^*d'{I^{-1}}^*\omega}_p$$ Thus $$ d\omega =I^*d'{I^{-1}}^*\omega =I^*{I^{-1}}^*d'\omega =d'\omega $$ $\blacksquare$

Proof. Let $(\varphi,U)$ be a smooth chart of $M$ such that $p\in U$. Then $$ df_p\p{\eval{\pdv{}{x^i}}_p} =(\varphi^*d{\varphi^{-1}}^*f)_p\p{\eval{\pdv{}{x^i}}_p} =(d(f\circ\varphi^{-1}))_{\varphi(p)}\p{d\varphi_p\p{\eval{\pdv{}{x^i}}_p}} =\pdv{(f\circ\varphi^{-1})}{x^j}(\varphi(p))dx^j_{\varphi(p)}\p{d\varphi_p\p{\eval{\pdv{}{x^i}}_p}} =\eval{\pdv{}{x^j}}_p(f)dx^j_{\varphi(p)}\p{\eval{\pdv{}{x^i}}_{\varphi(p)}} =\eval{\pdv{}{x^i}}_p(f) $$ Thus $$ df_p(v) =df_p\p{v^i\eval{\pdv{}{x^i}}_p} =v^idf_p\p{\eval{\pdv{}{x^i}}_p} =v^i\eval{\pdv{}{x^i}}_p(f) =v(f) $$ $\blacksquare$

Proof. Let $p\in U$ and $v\in T_pU$, then $$ d(x^j\circ\varphi)_p(v) =v(x^j\circ\varphi) =v^j =\mu^j_p(v) $$ $\blacksquare$

Proof. $$ d(\omega_1+\omega_2)_p =\p{\varphi_p^*d{\varphi_p^{-1}}^*(\omega_1+\omega_2)}_p =\p{\varphi_p^*d{\varphi_p^{-1}}^*\omega_1+\varphi_p^*d{\varphi_p^{-1}}^*\omega_2}_p =d{\omega_1}_p+d{\omega_2}_p $$ $$ d(c\omega)_p =\p{\varphi_p^*d{\varphi_p^{-1}}^*(c\omega)}_p =\p{c\varphi_p^*d{\varphi_p^{-1}}^*\omega}_p =cd\omega_p $$ $$ d(\omega\wedge\eta)_p =\p{\varphi_p^*d{\varphi_p^{-1}}^*(\omega\wedge\eta)}_p =\p{(\varphi_p^*d{\varphi_p^{-1}}^*\omega)\wedge\eta+(-1)^k\omega\wedge(\varphi_p^*d{\varphi_p^{-1}}^*\eta)}_p =(d\omega)_p\wedge\eta_p+(-1)^k\omega_p\wedge(d\eta)_p $$ $$ d(d\omega)_p =\p{\varphi_p^*d{\varphi_p^{-1}}^*d\omega}_p =\p{\varphi_p^*d{\varphi_p^{-1}}^*\varphi_p^*d{\varphi_p^{-1}}^*\omega}_p =\p{\varphi_p^*d(d{\varphi_p^{-1}}^*\omega)}_p =\p{\varphi_p^*0}_p =0 $$ $\blacksquare$

Proof. Let $p\in M$, then there exist a smooth chart $(\varphi,U)$ of $M$ and a smooth chart $(\varphi,V)$ of $N$ such that $p\in U$ and $F(U)\subseteq V$. Transform $F$ and $\omega$ to $U$ and $V$, then $$ F^*d\omega =\varphi^*{\varphi^{-1}}^*F^*\psi^*d{\psi^{-1}}^*\omega =\varphi^*{(\psi\circ F\circ\varphi^{-1})}^*d{\psi^{-1}}^*\omega =\varphi^*d{(\psi\circ F\circ\varphi^{-1})}^*{\psi^{-1}}^*\omega =\varphi^*d{\varphi^{-1}}^*F^*\omega =dF^*\omega $$ Thus $(F^*d\omega)_p=(dF^*\omega)_p$. $\blacksquare$

Proof. Let $(\varphi,U)$ be a smooth chart of $M$, then for each $k$-tuple $J$ of $n$, $\omega(e_{J_1},\ldots,e_{J_k})$ is smooth on $U$. Let $I$ be a strictly increasing $(k-1)$-tuple of $n$. Then $$ (X\lrcorner\omega)^I =(X\lrcorner\omega)(e_{I_1},\ldots,e_{I_{k-1}}) =\omega(X,e_{I_1},\ldots,e_{I_{k-1}}) =\omega(X^ie_i,e_{I_1},\ldots,e_{I_{k-1}}) =X^i\omega(e_i,e_{I_1},\ldots,e_{I_{k-1}}) $$ which is smooth on $U$. Thus $X\lrcorner\omega$ is smooth. $\blacksquare$

Proof. Pick an arbitrary basis $E$ of $V$ and define $E'$ from $E$ by changing $E_1$ into $-E_1$, so that $E'_1=-E_1$. Then $[E]$ and $[E']$ are the desired equivalence classes. $\blacksquare$

Proof. If $n=0$, this is trivial. Now suppose $n\neq0$. Let $p\in M$, then there exists a local frame $(F_1,\ldots,F_n)$ of $p$ on $U$, such that $\mathcal O|_UO(F_1,\ldots,F_n)=1$. Note that $(-F_1,F_2,\ldots,F_n)$ is a frame on $U$. And $(-\mathcal O)|_UO(-F_1,F_2,\ldots,F_n)=1$. Thus $-\mathcal O$ is an orientation of $M$. $\blacksquare$

Proof. Clearly, $\{f\omega:f\in C^\infty(M)\}\subseteq\Omega^n(M)$. Now let $\psi\in\Omega^n(M)$. Let $p\in M$, since $\Lambda^n(T^*_pM)$ has dimensionality $1$, and $\omega_p$ is non-zero, there exists $c_p\in R$ such that $c_p\omega_p=\psi_p$. We can define a function $f:M\to R$ such that $f(p)=c_p$. Then $f\omega=\psi$. Let $(\varphi,U)$ be a smooth chart of $M$, then we still have $f\omega=\psi$ when transformed to $U$. Thus $f\omega_Ie^I=\psi_Ie^I$. Then $f\omega_I=\psi_I$. Since $\omega_I$ is nowhere-zero, $f=\frac{\psi_I}{\omega_I}$, which is smooth. Since $f$ is smooth when restricted on every smooth chart of $M$, $f$ is smooth. $\blacksquare$

Proof. The goal is to show that $\{e\in\mathcal B_V|\omega(e_1,\ldots,e_n)\gt0\}=[b]$ for some $b\in\mathcal B_V$. Let $a=(a_1,\ldots,a_n)$ be a basis of $V$, then $\omega(a_1,\ldots,a_n)\neq0$. Define $b$ to be $a$ if $\omega(a_1,\ldots,a_n)\gt0$ and $(-a_1,a_2,\ldots,a_n)$ if $\omega(a_1,\ldots,a_n)\lt0$, then $b$ is a basis of $V$ with $\omega(b_1,\ldots,b_n)\gt0$. Let $e$ be any basis of $V$. Let $e^I$ be the basis of $\Lambda^n(V^*)$ defined with $e$, and let $A$ denote the transition matrix from $b$ to $e$, then $$\omega(b_1,\ldots,b_n) =\omega_Ie^I(b_1,\ldots,b_n) =\omega_I\det\p{\p{\mu^i(b_j)}_{ij}} =\omega_I\det(A) =\det(A)\omega_Ie^I(e_1,\ldots,e_n) =\det(A)\omega(e_1,\ldots,e_n)$$ Thus $\omega(e_1,\ldots,e_n)\gt0$ if and only if $\det(A)\gt0$. $\blacksquare$

Proof. If $n=0$, this is trivial. Now suppose $n\neq0$.

Suppose $\omega\in\Omega^n(M)$ is nowhere-zero. Let $p\in M$ and let $(\varphi,U)$ be a smooth chart of $M$ such that $p\in U$ and $U$ is a regular (half) ball. Then $(\pdv{}{x^1},\ldots,\pdv{}{x^n})$ is a local frame of $p$ on $U$. Note that $\omega=fdx^I$ for some smooth $f$ on $U$. Since $\omega$ is nowhere-zero, $f$ is also nowhere-zero. Since $U$ is a regular (half) ball, it is path-connected, and by intermediate value theorem, $f$ is alway positive or always negative. Note that $\omega_q(\pdv{}{x^1}_q,\ldots,\pdv{}{x^n}_q)=f(q)$ for all $q\in U$. If $f$ is always positive, then $\mathcal O_\omega|_UO(\pdv{}{x^1},\ldots,\pdv{}{x^n})=1$; If $f$ is always negative, then $\mathcal O_\omega|_UO(-\pdv{}{x^1},\pdv{}{x^2},\ldots,\pdv{}{x^n})=1$. Thus $\mathcal O_\omega$ is an orientation of $M$.

Now suppose $\mathcal O$ is an orientation of $M$. Let $p\in M$, then we can find a smooth chart $(\varphi_p,U_p)$ of $M$ such that $p\in U_p$ and an oriented frame $({F_p}_1,\ldots,{F_p}_n)$ on $U_p$. Let $({G_p}^1,\ldots,{G_p}^n)$ be the dual coframe of $({F_p}_1,\ldots,{F_p}_n)$. Then $\bigwedge {G_p}^j$ is a smooth section of $\Lambda^nT^*M$ on $U_p$. Let $\omega^p$ denote the extension of $\bigwedge {G_p}^j$ to $M$ by filling $M\setminus U$ with zeros. Note that $(U_p)_{p\in M}$ is an open cover of $M$. Let $(\psi_p)_{p\in M}$ be a smooth partition of unity with respect to $(U_p)_{p\in M}$. Then for each $p\in M$, $\psi_p\omega^p$ is an $n$-form of $M$. Let $\omega$ denote $\sum\psi_p\omega^p$ where for each $q\in M$, the summation sums over $p\in M$ such that $\psi_p(q)\neq0$. Then $\omega$ is an $n$-form of $M$. Let $q\in M$, then there exists $p\in M$ such that $\psi_p(q)\neq0$. Let $p$ be such, then $q\in U_p$. Note that the bases $({{F_p}^1}_q,\ldots,{{F_p}^n}_q)$ and $({{F_q}^1}_q,\ldots,{{F_q}^n}_q)$ of $T_pM$ both belong to the orientation $\mathcal O_p$ of $T_qM$. Thus $\omega^p_q({{F_p}^1}_q,\ldots,{{F_p}^n}_q)$ and $\omega^p_q({{F_q}^1}_q,\ldots,{{F_q}^n}_q)$ share the same sign. Since $\omega^p_q({{F_p}^1}_q,\ldots,{{F_p}^n}_q)=\bigwedge {{G_p}^j}_q({{F_p}^1}_q,\ldots,{{F_p}^n}_q)=1$, $\omega^p_q({{F_q}^1}_q,\ldots,{{F_q}^n}_q)\gt0$. Hence $\omega_q({{F_q}^1}_q,\ldots,{{F_q}^n}_q)\gt0$. We have shown that $\omega$ is nowhere-zero, thus $\mathcal O_\omega$ is an orientation of $M$. Let $q\in M$, then $({{F_q}^1}_q,\ldots,{{F_q}^n}_q)\in O_{\omega_q}$, implying $\mathcal O_\omega(q)=O_{T_qM}({{F_q}^1}_q,\ldots,{{F_q}^n}_q)=\mathcal O(q)$. Therefore, $\mathcal O_\omega=\mathcal O$. $\blacksquare$

Proof. If $n=0$, then $M$ contains exactly one point, and the proposition trivially follows. Now we assume $n\neq0$. Since $M$ is orientable, there exists an orientation $\mathcal O$ of $M$. Then $-\mathcal O$ is also an orientation of $M$. Since $M$ is non-empty, $\mathcal O\neq-\mathcal O$. Let $\mathcal O'$ be an orientation of $M$. Then for some $\omega,\omega'\in\Omega^n(M)$ that are nowhere-zero, $\mathcal O_\omega=\mathcal O$ and $\mathcal O_{\omega'}=\mathcal O'$. Thus for some $f\in C^\infty(M)$ that is nowhere-zero, $f\omega=\omega'$. Since $M$ is connected, $f(M)$ is connected, thus $f$ is either always positive or always negative. Then $\mathcal O_\omega$ and $\mathcal O_{\omega'}$ are either globally identical or globally opposite. Thus either $\mathcal O'=\mathcal O$ or $\mathcal O'=-\mathcal O$. $\blacksquare$

Proof. There exists $E\in O$. Let $E'$ denote $({dI_p}^{-1}(E_1),\ldots,{dI_p}^{-1}(E_n))$, then $E'$ is a basis of $T_pU$. The goal is to show that $[E']=\{B':B\in O\}$. Let $(\varphi,V)$ be a smooth chart of $U$ such that $p\in V$. Let $B$ be a basis of $T_pM$, $A$ the transition matrix from $B$ to $(\eval{\pdv{}{x_i}}_p^M)$, and $A'$ the transition matrix from $B'$ to $(\eval{\pdv{}{x_i}}_p^U)$. Let $\phi$ denote the coordinate map of $(\eval{\pdv{}{x_i}}_p^M)$ and $\phi'$ that of $(\eval{\pdv{}{x_i}}_p^U)$. Then $A_{*,j}=\phi(B_j)=\phi'(B'_j)=A'_{*,j}$, implying $A=A'$. If $F\in[E']$, then there exists a basis $B$ of $T_pM$, such that $B'=F$. Since $B'\sim E'$, $B\sim E$, implying $B\in O$. Thus $F\in\{B':B\in O\}$. Now suppose $F\in\{B':B\in O\}$, then for some $B\in O$, $B'=F$. Since $B\sim E$, $B'\sim E'$, implying $F\in[E']$. Hence $[E']=\{B':B\in O\}$. $\blacksquare$

Proof. Let $p\in U$, then there exists a local frame $(F_1,\ldots,F_n)$ of $p$ on an open subset $S$ of $M$ containing $p$. Define $G_j$ on $U\cap S$ by ${G_j}_q={dI_q}^{-1}{F_j}_q$ for all $q\in U\cap S$. Since ${dI_q}^{-1}$ is an isomorphism, $({G_1}_q,\ldots,{G_n}_q)$ is a basis. Since ${dI_q}^{-1}$ preserves coordinates under standard bases of any smooth chart, $G_j$ is smooth. Then $(G_1,\ldots,G_n)$ is a local frame of $p$, and $O(G_1,\ldots,G_n)$ is by definition equivalent to $\mathcal O^U|_{U\cap S}$. Thus $\mathcal O^U$ is an orientation of $U$.

Suppose $\omega\in\Omega^n(M)$ is nowhere-zero, then $I^*\omega$ is by definition nowhere-zero. Let $p\in U$, then ${{\mathcal O_\omega}^U}_p$ is the set of bases $({dI_p}^{-1}(B_1),\ldots,{dI_p}^{-1}(B_n))$ where $B$ is a basis of $T_pM$ such that $\omega_p(B_1,\ldots,B_n)\gt0$ and ${\mathcal O_{I^*\omega}}_p$ is the set of bases $B'$ of $T_pU$ such that $\omega_p(dI_p(B'_1),\ldots,dI_p(B'_n))\gt0$. Trivially, ${{\mathcal O_\omega}^U}_p={\mathcal O_{I^*\omega}}_p$. $\blacksquare$

Proof. For each $p\in M$, let $(\varphi_p,U_p)$ be a smooth chart of $M$ such that $p\in U_p$. Let $(\psi_p)_{p\in M}$ be a smooth partition of unity with respect to $(U_p)_{p\in M}$. For each $p\in M$, define the section $X^p:M\to TM$ by $X^p_q=-\psi_p(q)\eval{\pdv{}{x_n}}_q$ if $q\in U_p$ and $X^p_q=0$ otherwise. Then $X=\sum_{p\in M}X^p$ is clearly a vector field on $M$ such that for all $p\in\partial M$, $X_p$ is outward-pointing. $\blacksquare$

Proof. Suppose $(I^*(X\lrcorner\omega))_p=0$ for some $p\in\partial M$. Let $(\varphi,U)$ be a smooth chart of $M$ such that $p\in U$. Note that $dI_p(e_i^{\partial M})=e_i^M$ for all $i\in\{1,\ldots,n-1\}$. Thus $$ 0 =(I^*(X\lrcorner\omega))_p(e_1^{\partial M},\ldots,e_{n-1}^{\partial M}) =(X\lrcorner\omega)_p(e_1^M,\ldots,e_{n-1}^M) =\omega_p(X_p,e_1^M,\ldots,e_{n-1}^M) =X_p^n\omega_p(e_n^M,e_1^M,\ldots,e_{n-1}^M) $$ Note that $X_p^n\lt0$, thus $\omega_p(e_1^M,\ldots,e_n^M)=0$, implying $\omega_p=0$, a contradiction. Hence $I^*(X\lrcorner\omega)$ is nowhere-zero. $\blacksquare$

Proof. Suppose $\omega,\omega'$ are $n$-forms and $X,X'$ are vector fields as required. Let $p\in\partial M$ and let $(\varphi,U)$ be a smooth chart of $M$ such that $p\in U$. Then clearly, $B=(X_p,e_1^M,\ldots,e_{n-1}^M)$ and $B'=(X'_p,e_1^M,\ldots,e_{n-1}^M)$ are both bases of $T_pM$. Let $A$ be the transition matrix from $B$ to $B'$, then $$\det(A)=\varphi_{B'}(X_p)_1=\frac{X_p^n}{{X'}_p^n}\gt0$$ implying $\omega_p(X_p,e_1^M,\ldots,e_{n-1}^M)$ and $\omega_p(X'_p,e_1^M,\ldots,e_{n-1}^M)$ share the same sign. Since $\omega_p$ and $\omega'_p$ determine the same orientation of $T_pM$, $\omega_p(X'_p,e_1^M,\ldots,e_{n-1}^M)$ and $\omega'_p(X'_p,e_1^M,\ldots,e_{n-1}^M)$ share the same sign. Hence $(I^*(X\lrcorner\omega))_p(e_1^{\partial M},\ldots,e_{n-1}^{\partial M})$ and $(I^*(X'\lrcorner\omega'))_p(e_1^{\partial M},\ldots,e_{n-1}^{\partial M})$ share the same sign. We have shown that $I^*(X\lrcorner\omega)$ and $I^*(X'\lrcorner\omega')$ determine the same orientation of $\partial M$. $\blacksquare$

Proof. Since $F$ is a local diffeomorphism, there exist a smooth chart $(\varphi,U)$ of $M$ and a smooth chart $(\psi,V)$ of $N$, such that $p\in U$ and $F^U:U\to V$ is a diffeomorphism. Note that $$dF_p=d(I^V)_{F(p)}\circ d(F^U)_p\circ(d(I^U)_p)^{-1}$$ Since $d(I^V)_{F(p)},d(F^U)_p,(d(I^U)_p)^{-1}$ are isomorphisms, $dF_p$ is an isomorphism. $\blacksquare$

Proof. Let $p\in M$. Let $(e_i)$ be a basis of $T_pM$. Then $$(F^*\omega)_p(e_1,\ldots,e_k)=\omega_{F(p)}(dF_p(e_1),\ldots,dF_p(e_k))$$ Since $dF_p$ is an isomorphism, $(dF_p(e_i))$ is a basis of $T_{F(p)}N$. Thus $(F^*\omega)_p=0$ if and only if $\omega_{F(p)}=0$. Since $\omega$ is nowhere-zero, $F^*\omega$ is also nowhere-zero. $\blacksquare$

Proof. If $n=0$, this is trivial. Now suppose $n\neq0$. Let $\mathcal O^M$ be the induced orientation by $N$ and $F$. Let $p\in M$ and $B$ a basis of $T_pM$ such that $O_{T_pM}(B)=\mathcal O^M(p)$. Then $B\in O_{(F^*\omega)_p}$, implying $(F^*\omega)_p(B)\gt0$. Thus $\omega_{F(p)}(dF_p(B))\gt0$, implying $dF_p(B)\in O_{\omega_{F(p)}}$. Hence $O_{T_{F(p)}N}(dF_p(B))=\mathcal O^N(F(p))$. We have shown that $F$ is orientation-preserving with $\mathcal O^M$. Conversely, if $B$ is a basis of $T_pM$ such that $O_{T_{F(p)}N}(dF_p(B))=\mathcal O^N(F(p))$, then $O_{T_pM}(B)=\mathcal O^M(p)$. Now suppose we have an orientation ${\mathcal O^M}'$ of $M$ such that $F$ is orientation-preserving. Then given $p\in M$, there exists a basis $B$ of $T_pM$ such that $O_{T_pM}(B)={\mathcal O^M}'(p)$, implying $O_{T_{F(p)}N}(dF_p(B))=\mathcal O^N(F(p))$, and thus $O_{T_pM}(B)=\mathcal O^M(p)$. Hence ${\mathcal O^M}'=\mathcal O^M$. We have shown that $\mathcal O^M$ is the unique orientation of $M$ such that $F$ is orientation-preserving. Thus this induced orientation is independent on the choice of $\omega$. $\blacksquare$

Proof. Let $\omega$ be an orientation form of $M$. Note that $M$ has a countable basis $(\varphi_i,U_i)$ of regular (half) balls, which are non-empty. Fix $(\varphi_i,U_i)$ for some $i$. Define $f(q)=\omega_q({e_q}_1,\ldots,{e_q}_n)$ on $U_i$, which is nowhere-zero. Since $U_i$ is a regular (half) ball, it is path-connected, and by intermediate value theorem, $f$ is alway positive or always negative. Thus $(\varphi_i,U_i)$ is either oriented or anti-oriented.

Since $(U_i)$ covers $M$, it has a smooth partition of unity $(\psi_i)$. Let $i$ be an index. Since $U_i$ is a regular (half) ball, $\overline{U_i}$ is compact. Since $\text{supp}\psi_i$ is a closed subset of $\overline{U_i}$, it is compact. Thus each $\psi_i$ is compactly supported. $\blacksquare$

Proof. Note that $(\varphi,U\cap V)$ (as a chart of $U\cap V$) inherits the orientedness or anti-orientedness of $(\varphi,U)$, and $(\psi,U\cap V)$ (as a chart of $U\cap V$) inherits the orientedness or anti-orientedness of $(\psi,V)$. Let $\hat p\in\varphi(U\cap V)$. Then there exists $p\in U\cap V$ such that $\varphi(p)=\hat p$. Let $A$ denote the transition matrix from $B_\varphi=(e_1,\ldots,e_n)$ to $B_\psi=(e'_1,\ldots,e'_n)$ for $T_p(U\cap N)$. Then $$ A_{ji} =(e_i)^j_\psi =e_i(x^j\circ\psi) =\pdv{(\psi\circ\varphi^{-1})^j}{x^i}(\hat p) =(D(\psi\circ\varphi^{-1})(\hat p))_{ji} $$ Thus $$ \det(D(\psi\circ\varphi^{-1}))(\hat p) =\det(D(\psi\circ\varphi^{-1})(\hat p)) =\det(A) $$ If $(\varphi,U)$ and $(\psi,V)$ are both oriented or both anti-oriented, then $B_\varphi$ and $B_\psi$ belong to the same orientation, implying $\det(A)\gt0$. If one of $(\varphi,U)$ and $(\psi,V)$ is oriented and the other anti-oriented, then $B_\varphi$ and $B_\psi$ belong to opposite orientations, implying $\det(A)\lt0$. $\blacksquare$

Proof. Let $(p_i)$ and $(p'_j)$ be indexes of $M$, and suppose $$\sum\abs{\omega_{p_i}}$$ converges. Since $\abs{\omega_p}$ is unsigned for each $p\in M$, we have $$ \sum\abs{\omega_{p'_j}} =\sum_{p\in M}\abs{\omega_p} =\sum\abs{\omega_{p_i}} $$ We have shown that the integrability of $\omega$ is independent on the choice of index on $M$.

Suppose $\omega$ is integrable. Then given any index $(p_i)$ of $M$, $\sum\abs{\mathcal O(p_i)\omega_{p_i}}$ converges, thus $\sum\mathcal O(p_i)\omega_{p_i}$ converges.

Suppose $\omega$ is integrable and let $(p_i)$ and $(p'_j)$ be indexes of $M$. Since $\sum\abs{\mathcal O(p_i)\omega_{p_i}}$ converges, and $(p'_j)$ is a reindex of $(p_i)$, we have $$\sum\mathcal O(p'_j)\omega_{p'_j}=\sum\mathcal O(p_i)\omega_{p_i}$$ Thus the integral is independent on the choice of index on $M$. $\blacksquare$

Proof. Let $(\varphi_i,U_i,\psi_i)$ and $(\varphi'_j,U'_j,\psi'_j)$ be charts and partitions as required, and suppose $$\sum_i\int_{\varphi_i(U_i)}\abs{(\varphi_i^{-1})^*(\psi_i\omega)}$$ converges. Recall that a continuous function on measurable domain is a measurable function, and if it is also unsigned, then it is integrable with unsigned integral. Then $$ \sum_j\int_{{\varphi'}_j(U'_j)}\abs{({\varphi'}_j^{-1})^*({\psi'}_j\omega)} =\sum_j\int_{{\varphi'}_j(U'_j)}\sum_i(\psi_i\circ{\varphi'}_j^{-1})\abs{({\psi'}_j\omega_J)\circ{\varphi'}_j^{-1}} =\sum_j\sum_i\int_{{\varphi'}_j(U'_j)}\abs{(\psi_i{\psi'}_j\omega_J)\circ{\varphi'}_j^{-1}} =\sum_j\sum_i\int_{{\varphi'}_j(U_i\cap U'_j)}\abs{(\psi_i{\psi'}_j\omega_J)\circ{\varphi'}_j^{-1}} /// =\sum_j\sum_i\int_{\varphi_i(U_i\cap U'_j)}\abs{\det(D({\varphi'}_j\circ\varphi_i^{-1}))((\psi_i{\psi'}_j\omega_J)\circ\varphi_i^{-1})} $$ Note that $$ \det(D({\varphi'}_j\circ\varphi_i^{-1}))((\psi_i{\psi'}_j\omega_J)\circ\varphi_i^{-1}) =((\varphi_i^{-1})^*(\psi_i{\psi'}_j\omega))_I =(\psi_i{\psi'}_j\omega_I)\circ\varphi_i^{-1} $$ Thus $$ \sum_j\sum_i\int_{\varphi_i(U_i\cap U'_j)}\abs{\det(D({\varphi'}_j\circ\varphi_i^{-1}))((\psi_i{\psi'}_j\omega_J)\circ\varphi_i^{-1})} =\sum_j\sum_i\int_{\varphi_i(U_i\cap U'_j)}\abs{(\psi_i{\psi'}_j\omega_I)\circ\varphi_i^{-1}} =\sum_j\sum_i\int_{\varphi_i(U_i)}\abs{(\psi_i{\psi'}_j\omega_I)\circ\varphi_i^{-1}} =\sum_i\sum_j\int_{\varphi_i(U_i)}\abs{(\psi_i{\psi'}_j\omega_I)\circ\varphi_i^{-1}} =\sum_i\int_{\varphi_i(U_i)}\abs{(\varphi_i^{-1})^*(\psi_i\omega)} $$ Hence $$\sum_j\int_{\varphi'_j(U'_j)}\abs{({\varphi'}_j^{-1})^*(\psi'_j\omega)}$$ converges. We have shown that the integrability of $\omega$ is independent on the choice of smooth charts and partition of unity.

Suppose $\omega$ is integrable. Then given any charts and partitions $(\varphi_i,U_i,\psi_i)$ as required, $$\sum\int_{\varphi_i(U_i)}\abs{s(i)(\varphi_i^{-1})^*(\psi_i\omega)}$$ converges, implying $$\sum\abs{\int_{\varphi_i(U_i)}s(i)(\varphi_i^{-1})^*(\psi_i\omega)}$$ converges///, thus $$\sum\int_{\varphi_i(U_i)}s(i)(\varphi_i^{-1})^*(\psi_i\omega)$$ converges, and each term is real.

Suppose $\omega$ is integrable and let $(\varphi_i,U_i,\psi_i)$ and $(\varphi'_j,U'_j,\psi'_j)$ be charts and partitions as required. Then $$ \sum_j\int_{\varphi'_j(U'_j)}s'(j)({\varphi'}_j^{-1})^*(\psi'_j\omega) =\sum_j\int_{\varphi'_j(U'_j)}s'(j)\sum_i(\psi_i\circ{\varphi'}_j^{-1})(({\psi'}_j\omega_J)\circ{\varphi'}_j^{-1}) /// =\sum_j\sum_i\int_{\varphi'_j(U'_j)}s'(j)((\psi_i{\psi'}_j\omega_J)\circ{\varphi'}_j^{-1}) =\sum_j\sum_i\int_{\varphi'_j(U_i\cap U'_j)}s'(j)((\psi_i{\psi'}_j\omega_J)\circ{\varphi'}_j^{-1}) /// =\sum_j\sum_i\int_{\varphi_i(U_i\cap U'_j)}s'(j)\abs{\det(D({\varphi'}_j\circ\varphi_i^{-1}))}((\psi_i{\psi'}_j\omega_J)\circ\varphi_i^{-1}) $$ Note that $$ \abs{\det(D({\varphi'}_j\circ\varphi_i^{-1}))}=s(i)s'(j)\det(D({\varphi'}_j\circ\varphi_i^{-1})) $$ Thus $$ \sum_j\sum_i\int_{\varphi_i(U_i\cap U'_j)}s'(j)\abs{\det(D({\varphi'}_j\circ\varphi_i^{-1}))}((\psi_i{\psi'}_j\omega_J)\circ\varphi_i^{-1}) =\sum_j\sum_i\int_{\varphi_i(U_i\cap U'_j)}s(i)((\psi_i{\psi'}_j\omega_I)\circ\varphi_i^{-1}) /// =\sum_i\sum_j\int_{\varphi_i(U_i\cap U'_j)}s(i)((\psi_i{\psi'}_j\omega_I)\circ\varphi_i^{-1}) =\sum_i\sum_j\int_{\varphi_i(U_i)}s(i)((\psi_i{\psi'}_j\omega_I)\circ\varphi_i^{-1}) =\sum_i\int_{\varphi_i(U_i)}s(i)(\varphi_i^{-1})^*(\psi_i\omega) $$ We have shown that the integral is independent on the choice of smooth charts and partition of unity. $\blacksquare$