A paper I'm trying to understand uses the following lemma:

Let $p: U \to U_0$ be a topological covering map. Suppose that we can write $p =\pi \circ f$, where $f:U \to Y$ is an open surjective map, and $\pi: Y \to U_0$ is continuous. Then $f$ is also a topological covering map.

It seems like the proof should be very straightforward, but I can't get it to work out. I think I can prove it in the case of finite degree covering maps.

  • $\begingroup$ maybe sketch your proof for the finite degree map, and see where it might go wrong? $\endgroup$ – Henno Brandsma Feb 21 '12 at 21:32
  • $\begingroup$ @Henno Given a point $y\in Y$, consider an evenly covered nbd $N\subset Z$ containing $\pi(y)$. Suppose $N$ is evenly covered by $V_1,\ldots,V_n \subset V$. Then taking the intersection of all $g(V_i)\subset Y$ that contain $y$ gives an evenly covered nbd of $y$. $\endgroup$ – Ben Dozier Feb 21 '12 at 21:43

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.