Theorem numbers are those of the second edition.
This is the embedded submanifold with boundary version of Theorem 5.29.
Its title is "Restricting Maps to Submanifolds with Boundary" and the
tag for part (b) is "Restricting the Codomain". The statement is
that a smooth map $F$ from a smooth manifold with boundary $N$ into a smooth
manifold without boundary $M$ whose image is contained in an embedded submanifold
with boundary $S$ is smooth as a map into the submanifold with boundary.
The preceding theorem (Theorem 5.51) is the submanifold with boundary version
of Theorem 5.8 which connects slice conditions for submanifolds with boundary
to embedded submanifolds with boundary. I can prove Theorem 5.51 without trouble.
Just before the problematic theorem, Professor Lee writes "Using the preceding theorem in
place of Theorem 5.8, one can readily prove the following theorem." I have to
disagree with his use of "readily".
In my attempt, I followed the pattern of the proof of Theorem 5.8. We have a
$k$-dimensional embedded submanifold with boundary $S$ of the smooth n-dimensional manifold without boundary
$M$, so the $k$-slice condition for submanifolds with boundary applies and
by Theorem 5.51 we
get a slice chart $(W,\psi)$ for $S$ in $M$. Set $V_0=W\cap S$ and set
$\tilde{\psi}=\pi\circ\psi|_{V_0}$ where $\pi\colon\mathbb{R}^n\to\mathbb{R}^k$
is the projection on the first $k$ coordinates. If we can show that $(V_0,\tilde{\psi})$
is a smooth chart for $S$, then the rest of the proof goes through.
This part of the proof in Theorem 5.8 is contained in an errata entry (which
I believe itself has multiple typos). It shows that for any $q\in V_0$,
$d\tilde{\psi}_q$ is invertible. I wasn't thrilled with the proof in the errata, so
I came up with my own proof. Either way, I believe it to be true. Then comes
the part that doesn't translate properly from Theorem 5.8 to Theorem 5.53.
Professor Lee uses the Inverse Function Theorem for Manifolds (Theorem 4.5)
to show that $\tilde{\psi}$ restricts to a diffeomorphism; i.e., is a local
diffeomorphism. The problem is
that Theorem 4.5 only applies when the domain and codomain (modulo his comment that
the codomain restriction is not too severe if the map takes its values in
the interior) have empty boundaries. But in the case of Theorem 5.53, $V_0$
inherits a boundary from $S$. The proof of Theorem 5.8 then goes on to show
how $\tilde{\psi}$ must actually be a diffeomorphism, and that therefore
$(V_0,\tilde{\psi})$ is a smooth chart for $S$. It is obvious what the inverse of $\tilde{\psi}$
must be, but so far all my attempts at proving it smooth have failed.
Can somebody please suggest a way?
Best Answer
I found a way to finish the proof. Let $i\colon S\to M$ be the inclusion map, let $p\in N$ be arbitrary, and let $q=F(p)\in S$. Instead of taking an arbitrary slice chart at $q$ for $S$ in $M$ as promised by Theorem 5.51, we take the smooth chart for $S$ originally given in the proof of Theorem 5.51 (well, my proof, anyway). Call it $(U,\phi)$. The proof then gives the chart $(V,\psi)$ for $M$. Both of these charts are centered at $q$. The relationship between these charts is that $i(U)\subseteq V$, and for the smooth map $\hat{\imath}=\psi\circ i\circ\phi^{-1}\colon\phi(U)\to\psi(V)$, we have \begin{equation}\tag{1} \hat{\imath}(x^1,\dots,x^k)=(x^1,\dots,x^k,0,\dots,0) \qquad((x^1,\dots,x^k)\in\phi(U)). \end{equation} Next, the proof whittles down the two chart domains to $U_0$, a neighborhood of $q$ in $U$, and $V_0$, a neighborhood of $q$ in $V$, such that $U_0\subseteq V_0$. Since $U_0$ is open in $S$, there exists $W$ open in $M$ such that $U_0=W\cap S$. The proof then sets $V_1=V_0\cap W$. Then $U_0=U_0\cap V_0=S\cap W\cap V_0=V_1\cap S\subseteq V_1$, and $(V_1,\psi|_{V_1})$ is the slice chart centered at $q$ for $S$ in $M$ that we are going to use.
My proof for Theorem 5.53(b) then proceeds to define $\tilde{\psi}\colon U_0\to\pi(\psi(U_0))$ by $\tilde{\psi}=\pi\circ\psi|_{U_0}$ where $\pi\colon\mathbb{R}^n\to\mathbb{R}^k$ is the projection on the first $k$ coordinates.
For $q_0\in U_0$, say $\phi(q_0)=(x^1,\dots,x^k)$. Then \begin{equation} \psi(q_0) =\hat{\imath}(\phi(q_0)) =(x^1,\dots,x^k,0,\dots,0) \end{equation} by (1), and $$\tilde{\psi}(q_0)=\pi(\psi(q_0))=(x^1,\dots,x^k)=\phi(q_0),$$ hence $\phi|_{U_0}=\tilde{\psi}$. So $(U_0,\tilde{\psi}=\phi|_{U_0})$ is a smooth chart for $S$.
The rest of the proof is as for Theorem 5.29.