Restricting the codomain of a smooth map to submanifold is smooth

Question

Let $F:N\to M$ be a smooth map from a smooth manifold $N$ to a smooth manifold $M$ (without boundary) and $S$ an immersed submanifold of $M$ such that $F(N)\subset S$. If $F$ is continuous as a map from $N$ to $S$, then $F:N\to S$ is smooth.

This is theorem $5.29$ in John Lee's introduction to smooth manifolds $2$nd edition and in that same book he wrote the following proof (I will not give the full proof but only upto the part where I don't understand):

"$Proof$: Let $p\in N$ and $q=F(p)\in S$. Then as $S$ is an immersed submanifold, it is locally embedded. Therefore there is a neighborhood $V$ of $q$ such that $V$ is embedded in $M$, that is the inclusion $i|_V:V\hookrightarrow M$ is a smooth embedding. This implies $V$ satisfies the local slice criterion, that is there exist a smooth chart $(W, \psi)$ for $M$ that is a slice chart for $V$ centerd at $q$. The fact that $(W,\psi)$ is a slice chart means that $(V_0, \widetilde{\psi})$ is a smooth chart of $V$, where $V_0 = W\cap V$ and $\widetilde{\psi}= \pi\circ \psi|_{V_0}$, with $\pi: \mathbb{R}^n\to \mathbb{R}^k$ the projection onto the first $k=\dim S$ coordinates."

I understand everything upto this line. The next line below is where my confusion started.

"Since $V_0 = \left( i|_V\right)^{-1}(W)$ is open in $V$, it is open in $S$ in its given topology, and so $(V_0, \widetilde{\psi})$ is also a smooth chart for $S$."

I know that $V_0$ is open in $S$ and I know that $(V_0, \widetilde{\psi})$ will be a (continuous) chart but I dont understand the claim that $(V_0,\widetilde{\psi})$ is a smooth chart for S. How do we make sure that $(V_0, \widetilde{\psi})$ belongs to the smooth structure of $S$ by merely knowing $V_0$ is open? Is there a relationship between the smooth structure of $V$ and the smooth structure of $S$. Please help me I'm so confuse.

Answer 1

Personally I do not like the concept of an immersed submanifold as presented by Lee and other authors. The situation is this:

We have an injective immersion $j : P \to M$. This induces a unique topology and a unique smooth structure on the image set $S = j(P)$ such that $\bar j : P \stackrel{j}{\to} S^*$ becomes a diffeomorphism, where $S^*$ denotes the set $S$ endowed with the appropriate topology and smooth structure. This $S^*$ is called an immersed submanifold of $M$.

The problem here is that $j$ is in general no homeomorphism from $P$ onto its image $j(P)$ with the subspace topology inherited from $M$. Thus, although we call $S^*$ an immersed submanifold, it is in general not even a topological subspace of $M$. In my opinion this is somewhat confusing, but of course it is a matter of taste.

However, even if $S^*$ should not be a topological subspace of $M$ globally, it is always locally one. This means that each $q \in S^*$ has an open neigborhood $V$ in $S^*$ which is an emdedded submanifold of $M$. Note that $V$ is open in $S^*$, but in general not open in $S$ with the subspace topology inherited from $M$.

In my opinion a better formulation of Lee's Theorem 5.29 would be this:

Let $f :N \to P$ be a continuous function from a smooth manifold $N$ to a smooth manifold $P$ and let $j : P \to M$ be an immersion (which is not required to be injective). If $F = j \circ f$ is smooth, then $f$ is smooth.

Lee's theorem is a corollary. We are given an immersed submanifold $S^*$ of $M$ and a smooth $F : N \to M$ such that $F(N) \subset S^*$ and $\bar F : N \stackrel{F}{\to} S^*$ is continuous. This $S^*$ is associated to an injective immersion $j : P \to M$ such that $\bar j : P \stackrel{j}{\to} S^*$ is a diffeomorphism. The function $f = (\bar j)^{-1} \circ \bar F : N \to P$ is continuous and $j \circ f = F$ is smooth, thus $f$ is smooth. Hence $\bar F = \bar j \circ f$ is smooth.

Our above theorem can be proved exactly as Lee's Theorem 5.29. Let us now see why $(V_0, \widetilde{\psi})$ is a smooth chart for $S^*$.

Since $V$ is open in $S^*$ and $V$ is an embedded submanifold of $M$, it suffices to show that $(V_0, \widetilde{\psi})$ is a smooth chart for $V$. But this follows from a theorem in Lee's book which says that each subspace $E \subset M$ which is an embedded submanifold has a unique smooth structure such that the inclusion map $E \to M$ is a smooth embedding. A smooth atlas on $E$ generating this smooth structure is given precisely by the collection of all $(W_0 = W \cap E, \pi \circ \psi \mid_{W_0})$ where the $(W,\psi)$ are smooth charts for $M$ which are slice charts for $E$ centered at the points $q \in E$.

Perhaps it is somewhat confusing that the smooth structure on $V$ is primarily induced by $j_V : j^{-1}(V) \stackrel{j}{\to} V$. However, since we know that $V$ is an embedded submanifold of $M$, this smooth structure agrees with the above smooth structure induced by slice charts.

Answer 2

Due to a problem pointed out in the comments, it seems to me that the proof given in the text is not so easy to fix. The chart obtained by restricting the slice chart for $V$ may not be smoothly compatible with the original smooth structure on $V$; the only reason we know it is compatible is due to the uniqueness theorem (Theorem 5.31), but this result is proven using Theorem 5.29!

But here is a proof of Theorem 5.29 that avoids the issues and does not use slice charts.

Let $F_0: N \to S$ denote the map $F$ with its codomain restricted to $S$ (with its intrinsic smooth structure). Then we have $F = \iota \circ F_0$ with $\iota: S \hookrightarrow M$ the inclusion map. Let $\dim M = m$, $\dim N = n$ and $\dim S = k$. Also let $p \in N$ be arbitrary.

By hypothesis, $\iota$ is a smooth immersion between smooth manifolds, so the Constant Rank Theorem shows that there exist smooth charts $(U, \varphi)$ for $S$ such that $F(p) \in U$ and $(V, \psi)$ for $M$ such that $\iota(U) \subseteq V$ and with respect to which $\iota$ has the coordinate representation $\widehat{\iota} = \psi \circ \iota \circ \varphi^{-1}$: $$\widehat{\iota}(x^1, ..., x^k) = (x^1, ..., x^k, 0, ..., 0)$$

Since $F_0$ is continuous and $U$ is open in $S$, the set $F_0^{-1}(U)$ is open in $N$. Let $(W, \tau)$ be a smooth chart for $N$ such that $p \in W \subseteq F_0^{-1}(U)$.

Let $\widehat{F} = \psi \circ F \circ \tau^{-1}$ be the coordinate representation of $F$ with respect to the smooth charts $(W, \tau)$ and $(V, \psi)$, and let $\widehat{F_0} = \varphi \circ F_0 \circ \tau^{-1}$ be the coordinate representation of $F_0$ with respect to $(W, \tau)$ and $(U, \varphi)$. Then we have $$\widehat{F} = \psi \circ F \circ \tau^{-1} = \psi \circ (\iota \circ F_0) \circ \tau^{-1} = (\psi \circ \iota \circ \varphi^{-1}) \circ (\varphi \circ F_0 \circ \tau^{-1}) = \widehat{\iota} \circ \widehat{F_0}$$

Let $\pi_i: \mathbb{R}^n \to \mathbb{R}$ be the projections onto the $i$-th coordinate for $i = 1, \ldots, n$ and also let $p_i: \mathbb{R}^k \to \mathbb{R}$ again be projections onto the $i$-th coordinate (with the domain being $\mathbb{R}^k$ instead).

By the above expression for $\widehat{\iota}$, it is clear that $\pi_i \circ \widehat{\iota} = p_i$ for each $i = 1, \ldots, k$. Then for each $i$ we have $$p_i \circ \widehat{F_0} = \pi_i \circ \widehat{\iota} \circ \widehat{F_0} = \pi_i \circ \widehat{F}$$

Now for functions $\mathbb{R}^n \to \mathbb{R}^k$, smoothness is equivalent to smoothness of the component functions. Therefore, since $\widehat{F}$ is smooth, it follows that $\pi_i \circ \widehat{F}$ is smooth, and by the equality above, it follows that $p_i \circ \widehat{F_0}$ is smooth for each $i$. Therefore $\widehat{F_0}$ is smooth in the sense of calculus. This means that $(W, \tau)$ and $(U, \varphi)$ are smooth charts for $N$ and $S$ respectively such that $p \in W$, $F_0(W) \subseteq U$ and $\widehat{F_0}$ is smooth.

By definition of smoothness, this means $F_0$ is smooth.