Let $M,N,S$ be smooth manifolds with boundary such that $S\subset M$ is an embedded submanifold. Problem 9-13 from J. M. Lee, Introduction to Smooth Manifolds, 2nd ed., asks in particular to prove the following result:
Let $F:N\to M$ be a smooth map such that $F(N)\subset S$. Show that $F_S:N\to S$, the restriction of $F$ to $S$ in the codomain, is also smooth.
If it were $\partial M=\varnothing$, then I think know how to prove this: We would need to take a local half-slice chart $(U,\varphi)$ of $M$ for $S$; in turn, we get an induced local chart $(U\cap S,\tilde{\varphi})$ of $S$ (this is what one does in the proof of Theorem 5.51 of the same book). Assuming that $\varphi(U)$ is a ball, we can project $\varphi(U)$ into the local coordinates of $S$ to get a smooth map $\pi:\varphi(U)\to\tilde{\varphi}(U\cap S)$. If $\iota:S\hookrightarrow M$ is the inclusion, then the map $\pi\circ\varphi\circ F|_{F^{-1}(U)}\circ\iota|_{F^{-1}(U\cap S)}=\tilde{\varphi}\circ F_S|_{F_S^{-1}(U\cap S)}$ is smooth. Hence, $F_S|_{F_S^{-1}(U\cap S)}$ is smooth.
Now suppose $\partial M\neq\varnothing$ and let $p\in N$. To show that $F_S$ is smooth at $p$, the case $F(p)\in\operatorname{Int}M$ reduces to the previous situation. On the other hand, the case $F(p)\in\partial M$ is more complicated, since we do not have a local (half-)slice condition for submanifolds with or without boundary of manifolds with boundary. This is because a submanifold with or without boundary of a manifold with boundary $M$ can intersect $\partial M$ in complicated ways.
So I have two questions:
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Is the sketch above for the case $\partial M=\varnothing$ correct? I haven't checked the details but I think the idea works (tell me if there is anything relevant I'm forgetting). (This case was already asked here.)
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How one would deal with the case $\partial M \neq\varnothing$? Note that the only thing we did in the case $\partial M =\varnothing$ was to prove that $S$, inside $M$, is locally a retraction. I.e., that each point of $S$ has an open neighborhood $U\subset M$ such that there is a smooth retraction from $U$ onto $U\cap S$ (and then postcomposing this smooth retraction with $F|_{F^{-1}(U)}$ while precomposing with the inclusion to obtain that $F_S|_{F_S^{-1}(U\cap S)}$ is smooth). Maybe we should try to show that $S$ is still locally a retraction in the case $\partial M\neq\varnothing$? The only idea that comes to my mind to show this is by trying to classify all the possible ways in which $S$ can intersect $\partial M$, and then constructing an ad-hoc retraction for each situation. However, I don't think this classification is an easy thing to accomplish.
Best Answer
For the case where $M$ is a smooth manifold with boundary and $S$ is an embedded submanifold with boundary, the idea is to embed $M$ into the double of $M$, which is a smooth manifold with boundary so the previous result applies.
Let $\iota: S \to M$ be the inclusion map.
Let $N$ be a smooth manifold with or without boundary.
Suppose that $F: N \to M$ is a smooth map whose image is contained in $S$ and $F$ is continuous as a map from $N$ to $S$. Let $F_S: N \to S$ be this restricted map.
By Example 9.32 and Theorem 9.29 in the book, $M$ can be smoothly embedded into the double $D(M)$, which is a smooth manifold without boundary. Let $e: M \to D(M)$ be the smooth embedding.
Since $S$ is an embedded submanifold with boundary of $M$, the inclusion map $\iota: S \to M$ is a smooth embedding. Then $e \circ \iota: S \to D(M)$ is a smooth embedding. Let $S' = e(S)$, then $e \circ \iota$ restricts to a diffeomorphism onto $S'$ given by $e_S: S \to S'$. Then $\iota' \circ e_S = e \circ \iota$, so $\iota' = (e \circ \iota) \circ (e_S)^{-1}$ is a smooth embedding. Then $S'$ is an embedded submanifold with boundary of $D(M)$.
Now $e_S \circ F_S$ is a continuous map from $N$ to $S'$, and $e \circ F$ is a smooth map from $N$ to $D(M)$. Moreover, $\iota’ \circ (e_S \circ F_S) = (\iota’ \circ e_S) \circ F_S = (e \circ \iota) \circ F_S = e \circ (\iota \circ F_S) = e \circ F$, so the image of $e \circ F$ lies in $S’$ and $e \circ F$ restricts to a continuous map onto $S'$.
But $D(M)$ is a smooth manifold without boundary, so we can apply Theorem 5.53 (b) to show that $e_S \circ F_S$ is smooth. Since $e_S$ is a diffeomorphism, it follows that $F_S = (e_S)^{-1} \circ (e_S \circ F_S)$ is smooth by composition.