If a Subset Admits a Smooth Structure Which Makes it into a Submanifold, Then it is a Unique One.

Question

$$ \newcommand{\wh}{\widehat} \newcommand{\R}{\mathbf R} \newcommand{\mr}{\mathscr} \newcommand{\set}[1]{\{#1\}} \newcommand{\inclusion}{\hookrightarrow} \newcommand{\vp}{\varphi} $$

I am trying to understand the following theorem:

Theorem. Let $M$ be a smooth manifold and $S$ be a subset of $M$. There is a unique smooth structure on $S$, if one exists, which makes it into a smooth manifold such that $i:S\inclusion M$ is a smooth embedding.

The only way I could prove this is via the lemma proved below whose proof is very long. I think the proof of the above theorem should be fairly straightforward and clear and should not require to do what I have done. Does somebody have a short proof?

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Lemma. Let $S$ be a $k$-dimensional embedded submanifold of a smooth manifold $M$. Then for each $p\in S$, there exist a smooth charts $(U_p, \vp_p)$ and $(V_p, \psi_p)$ about $p$ on $S$ and $M$ respectively such that $U_p=V_p\cap S$, and $\psi_p\circ\vp_p^{-1}:\wh U_p\to \wh V_p$ (the `hat' denotes the image of the open set under the corresponding chart) \begin{equation*} \psi_p\circ\vp_p^{-1}(x_1 , \ldots, x_k) = (x_1 , \ldots, x_k, 0 , \ldots, 0) \end{equation*} for all $(x_1 , \ldots, x_k)\in \wh U_p$. Therefore $\vp_p=(\pi\circ \psi_p\circ i)|_{V_p\cap S}$, where $\pi:\R^n\to \R^k$ is the projection on the first $k$ coordinates, and the collection of smooth charts $\mr U=\set{V_p\cap S,\ (\pi\circ \psi_p\circ i)|_{V_p\cap S}}_{p\in S}$ is a smooth atlas on $S$.

Proof: Since $i:S\inclusion M$ is an immersion (it's more than that), by the Constant Rank Theorem, we know that there exists a smooth chart $(U,\vp)$ on $S$ containing the point $p$, and a smooth chart $(V,\psi)$ on $M$, again containing $p$, such that $U\subseteq V$ and $\psi\circ\vp^{-1}(x_1,\ldots,x_k)=(x_1,\ldots,x_k,0,\ldots,0)$ for all $(x_1,\ldots,x_k)\in \wh U$. Since $i:S\inclusion M$ is in particular a topological embedding, we know that $U$ is open in $M$ and thus we may WLOG assume that $U=V\cap S$.
We now show that $\vp=(\pi\circ \psi\circ i)|_{V\cap S}$. For take $q\in V\cap S$, and say $\vp(q)=(x_1 , \ldots, x_k)$. Now we have $\psi\circ \vp^{-1}(x_1 , \ldots, x_k)=(x_1, , \ldots, x_k, 0 , \ldots, 0)$, giving $(\pi\circ \psi\circ i)(q)=(x_1 , \ldots, x_k)$.

So we have shown that for each $p\in S$, there exist a smooth charts $(U_p, \vp_p)$ and $(V_p, \psi_p)$ about $p$ on $S$ and $M$ respectively such that $U_p=V_p\cap S$, and $\psi_p\circ\vp_p^{-1}:\wh U_p\to \wh V_p$ \begin{equation*} \psi_p\circ\vp_p^{-1}(x_1 , \ldots, x_k) = (x_1 , \ldots, x_k, 0 , \ldots, 0) \end{equation*} for all $(x_1 , \ldots, x_k)\in \wh U_p$. It remains to show that $\mr U=\set{U_p, \vp_p}_{p\in S}$ is a smooth atlas on $S$. To see this, consider note that \begin{equation*} \vp_q\circ \vp_p^{-1} = \vp_q\circ \psi_p^{-1} \circ \psi_p\circ \vp_p^{-1} \end{equation*} and this map sends $(x_1 , \ldots, x_k)$ to $\pi\circ\psi_q^{-1}\circ\psi_p(x_1 , \ldots, x_k, 0 , \ldots, 0)$ for all $(x_1 , \ldots, x_k)\in \vp_p(U_p\cap U_q)$, and hence is smooth.

Answer 1

I would be very surprised if you could avoid using some form of the lemma or the Constant Rank Theorem for maps $\mathbb{R}^k \to \mathbb{R}^n$. Just as we define smooth $n$-manifolds by relating them to $\mathbb{R}^n$, the property of being a smooth $k$-submanifold is inherently connected to the way $\mathbb{R}^k$ sits in $\mathbb{R}^n$. But we can replace the "lemma" with a special case of the Constant Rank Theorem: If $f: \mathbb{R}^k \to \mathbb{R}^n$ is smooth and $df_p$ is injective for some $p$, then there is a diffeomorphism $g: \mathbb{R}^n \to \mathbb{R}^n$ such that, near $p$, the composition $g \circ f$ is the canonical immersion $(x_1,\ldots,x_k) \mapsto (x_1,\ldots,x_k,0,\ldots,0)$.

Now for the uniqueness theorem:

Proof. We want to show that there is a unique maximal smooth atlas for $S$ such that the inclusion $i: S \hookrightarrow M$ is a smooth embedding. Therefore it suffices to show the following: If $\, \phi_1,\phi_2 : U \to \mathbb{R}^k$ are charts on $U \subset S$ such that $$i \circ \phi_1^{-1}, i \circ \phi_2^{-1}: \mathbb{R}^k \to M$$ are smooth, then $\phi_1$ and $\phi_2$ are compatible, i.e. $\phi_2 \circ \phi_1^{-1}$ is smooth. To show this, fix a chart $\psi: V\subset M \to \mathbb{R}^n$ where $V \cap S=U$. We know that $\psi \circ i \circ \phi_1^{-1}: \mathbb{R}^k \to \mathbb{R}^n$ is an immersion, so for any point $x \in\mathbb{R}^k$ there is a diffeomorphism $g: \mathbb{R}^n \to \mathbb{R}^n$ such that $g \circ( \psi \circ i \circ \phi_1^{-1})$ is the canonical immersion near $x$. Letting $\pi: \mathbb{R}^n \to \mathbb{R}^k$ be the canonical submersion, it follows that $$ \pi \circ g \circ \psi \circ i \circ \phi_1^{-1}=\operatorname{id}. \tag{1}$$ Finally, we have $$\phi_1 \circ \phi_2^{-1}= \operatorname{id} \circ (\phi_1 \circ \phi_2^{-1})=\pi \circ g \circ \psi \circ i \circ \phi_1^{-1} \circ (\phi_1 \circ \phi_2^{-1}) = \pi \circ g \circ \psi \circ i \circ \phi_1^{-1}. \tag{2}$$ The map on the right is smooth because $\pi$, $g$, $\psi$, and $i \circ \phi_1^{-1}$ are smooth. $\square$

Here's another perspective that uses the same ideas:

Lemma. Consider maps $X \overset{f}{\to} Y^k \overset{g}{\to} Z^n$ such that $f$ is continuous, $g$ is a smooth immersion, and $g\circ f$ is smooth. Then $f$ is also smooth.

Proof. Fix a point $x \in X$ and charts $(V,\phi)$ around $y=f(x) \in Y$ and $(W,\psi)$ around $z=g(y) \in Z$ in which $g$ is locally represented by the canonical immersion. Letting $\pi: \mathbb{R}^{n} \to \mathbb{R}^{k}$ be the canonical submersion, it follows that $\pi \circ \psi \circ g|_V \circ \phi^{-1}: \mathbb{R}^{k} \to \mathbb{R}^{k}$ is the identity. Since $f$ is continuous, we can find a neighborhood $U$ of $x$ such that $f(U)\subset V$. Then the composition $$ \phi \circ f|_U=\pi \circ \psi \circ g|_V \circ \phi^{-1} \circ \phi \circ f|_U =\pi \circ \psi \circ g|_V \circ f|_U$$ is smooth since $\pi$, $\psi$, and $g|_V \circ f|_U = (g \circ f)|_U$ are smooth. Since $\phi$ is a diffeomorphism, $f$ is smooth. $\square$

This also proves the uniqueness result you're after: If $i : S \hookrightarrow M$ is a smooth embedding and you have another smooth embedding $i': S' \to M$ with $i'(S')=i(S)$, then there is an obvious continuous bijection $f:S' \to S$ such that $i\circ f=i'$. The preceding lemma implies that $f$ and $f^{-1}$ are smooth, hence $S'$ and $S$ are diffeomorphic.