Prove that if $f:X\to Y$ is a diffeomorphism of manifolds with boundary, then $f$ maps $\partial X$ to $\partial Y$ diffeomorphically.
Answer:
Let $U\subset H^k$ be an open subset and let $\phi:U\rightarrow X$ be a parametrization of $X$, $f\circ \phi : U\rightarrow Y$ is a parametrization of $Y$. Then $\partial Y \cap f\circ \phi (U) = f\circ \phi (\partial U)$, thus $\partial Y \subset f(\partial X)$ as $Y$ is covered by such parametrizations. Similarly, $\partial X \subset f^{-1}(\partial Y)$ and thus $f(\partial X) = \partial Y$.
I don't understand why $\partial Y \cap f\circ \phi (U) = f\circ \phi (\partial U)$, I understand $f \circ \phi$ is a diffeomorphism but don't understand why it maps boundary of $U$ to boundary of $Y$. Thanks and appreciate a hint.
Best Answer
Here's an easier argument. If $f:X\to Y$ is a homeomorphism of manifolds with boundary, then it restricts to a homeomorphism of the boundaries. (Then if $f$ is a diffeomorphism, the restriction will also be a diffeomorphism.)
Proof:
There is a purely topological way to distinguish boundary points from non-boundary points.
Boundary points have contractible punctured neighborhoods, non-boundary points do not have contractible punctured neighborhoods. This is immediate since deleting the center of an open ball in $\Bbb{R}^k$ yields a space homotopic to the $k-1$-sphere, whereas deleting the center of an open ball that has been intersected with a half-space passing through the center gives a contractible space (it's star shaped around any point not on the boundary of the half-space).
Since this characterization of boundary points is purely topological, it is preserved by homeomorphisms.
How this addresses your question
Well, I hope it illuminates why $\partial Y\cap f \circ \phi(U) = f\circ \phi (\partial U)$. It's because homeomorphisms (and thus diffeomorphisms) preserve boundary points.