1
$\begingroup$

I don't understand a couple of things about the following proof:

Statement: Suppose $f:X\rightarrow Y$ is a diffeomorphism of manifolds with boundary. Show that $\partial f:\partial X\rightarrow \partial Y$ is a diffeomorphism.

Proof: First, let's verify that $f(\partial X)\subset \partial Y$ . Let $x \in \partial X$; $y = f(x) \notin \partial Y$ . Let $\psi : V\rightarrow U$ be a local parameterization of an open neighborhood of $y$ in $Y$ . Then $f^{-1}\circ \psi : V\rightarrow f^{-1}(U)$ is a diffeomorphism to an open neighborhood of $x$ in $X$. Hence $x$ is not in $\partial X$. Thus $f(\partial X) \subset \partial Y$ . By applying same arguments to $f^{-1}$ we conclude that $f(\partial X) = \partial Y$. Thus $\partial f : \partial X\rightarrow \partial Y$ is bijective (it is onto and injective since f is). $\partial f$ is smooth as the restriction of a smooth $f$ and its inverse is also smooth (again as the restriction of a smooth $f^{-1}$).

So I don't understand how $f^{-1}\circ \psi : V\rightarrow f^{-1}(U)$ is even defined because $\psi: V\rightarrow U$ and $f^{-1}:Y\rightarrow X$. Also, the conclusion that $x$ is not in $\partial X$ is arrived at because $x$ cannot be on the boundary and have an open neighborhood around it diffeomorphically mapped to an open nbd of $\partial Y$, am I correct? Thanks and appreciate a hint.

$\endgroup$
1
$\begingroup$

We have $U\subset Y$. Whenever $g:A\to B$ and $B\subset C$ and $h:C\to D$, when we write $h\circ g$, we really mean $h|_{B}\circ g$. And if $p:A\to C$ and $p(A) \subset B$, then we also often write $p$ for the function $p:A\to B$ which matches the other $p$ in the obvious way.

We are assuming $y\notin \partial Y$, so that $y$ has a neighborhood $U$ diffeomorphic to an open subset of $\mathbb{R}^n$ for some $n$. And $x$ has no such neighborhood, being a boundary point. But $f^{-1}(U)$ is such a neighborhood.

$\endgroup$
  • $\begingroup$ Thanks, I understand why $x$ is not in $\partial X$ now, but for the first paragraph, $\psi:V\rightarrow U$, $f^{-1}:Y\rightarrow X$ and $U\not\subset Y$, right? in fact $U\subset X$ which is confusing me? $\endgroup$ – manifolded Feb 28 at 16:35
  • $\begingroup$ No, $U \subset Y$ is a neighborhood of $y$ in $Y$ and $V \subset \mathbb{R}^n$ is open with $\psi:V\to U$ a chart for $U$. $\endgroup$ – csprun Feb 28 at 16:37
  • $\begingroup$ Grrr, yes, you are right! Thanks. $\endgroup$ – manifolded Feb 28 at 16:37

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.