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.


  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?


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.

  • $\begingroup$ 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

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