0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Interesting! May I know how you intuitively know that boundary is a contractible manifold or in general how do you look at an arbitrary manifold and know if it's contractible? I know just know the definition: $X$ is contractible if $id_X $ is homotopic to a constant map i.e. $X$ can be smoothly shrunk to a point. $\endgroup$ – manifolded Mar 17 at 19:03
  • $\begingroup$ Also, may I know what you mean by purely topological that makes it homeomorphism invariant? Thanks. $\endgroup$ – manifolded Mar 17 at 19:04
  • $\begingroup$ @manifolded The boundary is not contractible, at least not generally, but any point in the boundary has a contractible punctured open neighborhood. As I mentioned, take a sufficiently small open ball around the boundary point, which is homeomorphic to an ordinary open ball in $\Bbb{R}^n$ intersected with a half-space that passes through the center of the open ball. Deleting the center of the open ball leaves us with half an open ball minus the center point, and this remaining set is star shaped around any point not in the hyperplane that defined the half-space, and thus contractible. $\endgroup$ – jgon Mar 17 at 19:06
  • $\begingroup$ You may want to try proving from the definition that (nonempty) star shaped subsets of $\Bbb{R}^n$ are contractible. $\endgroup$ – jgon Mar 17 at 19:06
  • $\begingroup$ By purely topological, I mean that the existence of a contractible punctured open neighborhood of a point is something that is preserved by homeomorphisms. If this is not obvious, I suggest trying to prove it. $\endgroup$ – jgon Mar 17 at 19:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.