In Professor Lee's Introduction to Riemannian Manifolds, second edition on page 12,
the first paragraph on Isometries reads
Suppose $(M,g)$ and $(\tilde{M},\tilde{g})$ are Riemannian manifolds with or
without boundary. An isometry from $(M,g)$ to $(\tilde{M},\tilde{g})$ is a diffeomorphism $\phi\colon M\to\tilde{M}$ such that $\phi^*\tilde{g}=g$.
Unwinding the definitions shows that this is equivalent to the requirement that
$\phi$ be a smooth bijection and each $d\phi_p\colon T_pM\to T_{\phi(p)}\tilde{M}$
be a linear isometry.
Only one part of a proof eludes me. That is, given a smooth bijection $\phi$ such that each
$d\phi_p\colon T_pM\to T_{\phi(p)}\tilde{M}$ is a linear isometry, I haven't been able to prove that $\phi^{-1}$ is smooth if $M$ has nonempty boundary. When $M$
has no boundary, I can use his Introduction to Smooth Manifolds, second edition (ISM)
Proposition 4.8(a) and ISM Exercise 4.9 to get that $\phi$ is a local diffeomorphism
and ISM Proposition 4.6(f) to get that $\phi$ is a diffeomorphism. ISM Proposition 4.8(a)
relies on the Inverse Function Theorem for Manifolds (ISM Theorem 4.5) which
Professor Lee warns us "can fail for a map whose domain has nonempty boundary."
Is a map with invertible differential that maps boundary to boundary a local diffeomorphism? might be part of an answer if I
could prove that $\phi$ mapped the boundary of $M$ into the boundary of $\tilde{M}$. (I can already prove that it maps the interior of $M$ into the
interior of $\tilde{M}$ using ISM Problem 4-2.) Of course, such a requirement is necessary for $\phi$ to be a diffeomorphism by the
Diffeomorphism Invariance of the Boundary Theorem (ISM Theorem 2.18), but I haven't been able to prove that $\phi$ maps boundary to boundary either.
So, how can $\phi^{-1}$ be proven to be smooth, or is the statement in the book incorrect?
Best Answer
Given this asumptions, $\phi^{-1}$ cannot be proven to be smooth.
Consider $[0, 2\pi)$ and $S^1$ with standard smooth structures. Manifold $[0, 2\pi)$ has nonempty boundary $\partial [0, 2\pi) = \{0\}$.
The function $\phi\colon[0, 2 \pi) \to S^1$ given by $\phi(t) = (\cos(t), \sin(t))$ is smooth bijection, but not diffeomorphism - inverse function is not continuous at $(1, 0)$.
The part about $d\phi_t$ being linear isometry does not invalidate this example. This is kind of independent condition. But to be more formal, we can endow both spaces with metric tensors by setting $\frac\partial{\partial x^1}$ and $ d\phi\frac\partial{\partial x^1} $ to be one-element orthonormal basis in corresponding manifolds. Notice how vectors $d\phi_t \frac\partial{\partial x^1}$ are always nonzero, and thus definition is correct. This deifnition gives us that $d\phi_t\colon T_t [0, 2\pi) \to T_{\phi(t)}S^1$ is isometry for each $t$.