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Let $X,\ Y$ be manifolds in $\mathbb R^n$ (they are locally diffeomorphic to open subsets of some euclidean space.) I have a question about this fact (an exercise in Guillemin and Pollack):

An injective local diffeomorphism $f: X\rightarrow Y$ is a diffeomorphism onto an open subset of $Y$.

This seems too trivial to me and hence I think I musunderstand something. I would prove this claim as follows.

The map $f: X\rightarrow f(X)$ is bijective. It is differentiable at any point since it is locally smooth (and even locally diffeomorphic), and the inverse $f^{-1}: f(X)\rightarrow X$ is also differentiable at any point because given any point $f(x)\in f(X)$, the map $f$ is a diffeomorphism near $x$, so $f^{-1}$ is differentiable at $f(x)$. Thus $f$ is a global diffeomorphism. Because of this and since $X$ is open in itself, $f(X)$ is open in $Y$.

How bad/good does this proof look?

Ottavio
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user557
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  • You should be much more explicit in the step you conclude $f^{-1}$ is differentiable. How does the derivative of $f^{-1}$ relate with the derivative of $f$? – PVAL-inactive Jan 31 '18 at 21:54
  • @PVAL-inactive: I think the Inverse function theorem tells us how the derivative of $f^{-1}$ relate with that of $f$. But I still don't understand what is wrong with such a proof.

    Edit: or you mean that if $f'(x)=0$ then $f^{-1}$ is not differentiable at $f(x)$?

     – user557 Jan 31 '18 at 22:16
  • Correcting myself: $f'(x)$ cannot be the zero map: $f$ is a diffeomorphism at $x$. – user557 Jan 31 '18 at 22:35
  • I mean, it depends on your definition of a local diffeomorphism. By the inverse function theorem an equally good definition is that locally, the inverse is differentiable, and certainly this seems the more natural definition. In that case, your proof seems fine to me. – Ishan Levy Jan 31 '18 at 22:56
  • @IshanLevy I'm only aware of one definition (https://en.wikipedia.org/wiki/Local_diffeomorphism), and it certainly includes that locally the inverse is differentiable. – user557 Jan 31 '18 at 23:13
  • Right, and this definition is perfectly good, as is your proof. I think PVAL-inactive was using a different but equivalent definition, namely that the derivative is an isomorphism on each tangent space. – Ishan Levy Feb 01 '18 at 01:38
  • I think you need to add $X=\cup_{p \in X} U_p$ where $U_p$ is the set on which $f$ is diffeomorphic. Then $f(X)=\cup_{p \in X} f(U_p)$[As $f$ is injective]. Now we know $f(U_p)$ is open because of the wiki definition and hence $f(X)$ is open by the arbitrary union of open sets. Do you mean this or you have a shorter argument? – Ri-Li Jan 03 '21 at 22:25

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