2
$\begingroup$

Suppose that $f:M\rightarrow N$ is a continuous map with the property that $\forall x\in M\exists $ open neighbourhood $U\subset M$ with $x\in U$ and open neighbourhood $V\subset N$ with $f(x)\in V$ s.t. $f|:U\rightarrow V$ is a homemorphism. $M$ is compact and $N$ is connected and Hausdorff.

Then

  1. $f$ is surjective
  2. $f$ is finite-to-one
  3. $f$ is a covering map.

I'm trying to work out the first part. I know that $f(M)$ is compact and hence closed. Is $f(M)$ open? If so, why? And how does this imply surjectivity?

$\endgroup$
3
$\begingroup$

For every point $x\in M$ we may find an open $U_x\ni x$ such that $f|_{U_x}\colon U_x\to V_x$ is a homeomorphism for some open $V_x\subseteq N$. Hence $f(M) = f(\bigcup_{x\in M} U_x) = \bigcup_{x\in M} f(U_x) = \bigcup_{x\in M} V_x$, which is a union of open sets and thus open.

Now $f(M)\subseteq N$ is open and closed and thus by connectedness of $N$, $f(M)\in\{\varnothing,N\}$. That is, $M=\varnothing$ or $f$ is surjective.

$\endgroup$
  • $\begingroup$ More precisely, $f$ is an open mapping. $\endgroup$ – Stefan Hamcke Apr 13 '13 at 15:28
  • $\begingroup$ @StefanH. You're right. That is essentially the argument. $\endgroup$ – Abel Apr 13 '13 at 15:33

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.