# A question about diffeomorphism between manifolds with boundary

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.

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.
• 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? – manifolded Feb 28 at 16:35
• 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 $U$. – csprun Feb 28 at 16:37