1
$\begingroup$

We say that $f:f: M \to N$ is locally proper if for all $x \in M$ there exist $V \ni x $ open in $M$ such that $f|_{\overline{V}}:\overline{V} \to Y$ is proper.

I know two ways to prove it. First, just note that every map $f: M \to Y$ $C^{1}$ between manifolds $M$ e $N$ with dimension $m$ and $n$ respectively is a Fredholm Map. So it's a particular result that "Every Fredholm map is locally proper" to Banach Manifolds by Smale, 1965 in "An Infinite Dimensional Version of Sard's Theorem".

Second way, we can do it in this particular case:

Let $f: \mathbb{R}^{m} \to \mathbb{R}^{n}$ $C^{1}$ then $f$ is locally proper.

dem: Let $x \in\mathbb{R}^{m}$. We claim that $\tilde{f}:=f|_{B[x,\epsilon]}:B[x,\epsilon] \to \mathbb{R}^{n}$ is proper.

Let $K$ compact set in $\mathbb{R}^{n}$ so $K$ is closed. Since $f$ is continous $f^{-1}(K)$ is closed in $\mathbb{R}^{m}$. $f^{-1}(K)\cap B[x,\epsilon]$ is closed, but $B[x,\epsilon]$ is bounded, hence $f^{-1}(K)\cap B[x,\epsilon]$ is compact in $\mathbb{R}^{m}$.

But $\tilde{f}^{-1}(K)= f^{-1}(K)\cap B[x,\epsilon]$ is compact in $\mathbb{R}^{m}$ then $\tilde{f}$ is proper. We fineshed.

More generally, let $f: M \to Y$ $C^{1}$ map between manifolds $M$ e $N$ with dimension $m$ and $n$ respectively then $f$ is locally proper.

How can I do? Thanks!

$\endgroup$
  • $\begingroup$ $f^{-1}(K)\cap B(x,\epsilon)$ need not be compact (and indeed usually won't be) since it may not be closed in $\mathbb{R}^n$. $\endgroup$ – Eric Wofsey Apr 22 '16 at 17:37
  • $\begingroup$ Correct, I'll fix it $\endgroup$ – Alladin Apr 22 '16 at 22:36

Your Answer

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

Browse other questions tagged or ask your own question.