6
$\begingroup$

Let $f : X \to Y$ and $g : Y \to Z$ be continuous maps (between topological spaces). Assume these hypotheses:

  • $f : X \to Y$ is a split surjection, i.e. has a section.
  • $g \circ f : X \to Z$ is a local homeomorphism, i.e. there is an open cover $\{ U_i : i \in I \}$ of $X$ such that, for each $i \in I$, the composite $U_i \to X \to Y \to Z$ is an open embedding.

Does it follow that $g : Y \to Z$ is a local homeomorphism?


Here are some observations:

  • The question with "open map" instead of "local homeomorphism" has a positive answer. In particular, under the above hypotheses, $g : Y \to Z$ must be an open map.
  • Moreover, the fibres of $g : Y \to Z$ must be discrete. So we have an open map with discrete fibres – is such a thing necessarily a local homeomorphism?
  • If $f : X \to Y$ is an open map, then $g : Y \to Z$ is a local homeomorphism. Conversely, if $g : Y \to Z$ is a local homeomorphism, then $f : X \to Y$ is also a local homeomorphism (hence an open map a fortiori).
$\endgroup$

1 Answer 1

5
$\begingroup$

First, note that $f:X\to Y$ must be locally injective. Now choose a splitting of $f$ and consider $Y$ as a subspace of $X$ via this splitting. For any $y\in Y$, there then some neighborhood $U\subseteq X$ of $y$ on which $f$ is injective. But $f$ is the identity on $U\cap Y$, and so $f$ must map $U\cap (X\setminus Y)$ outside of $U\cap Y$. But then $U\cap f^{-1}(U\cap Y)$ is a neighborhood of $y$ in $X$ that is entirely contained in $Y$. This implies $Y$ is open in $X$, and it follows immediately that the restriction of $g$ to $Y$ is a local homeomorphism.

$\endgroup$
1
  • $\begingroup$ Thanks! For some reason I was sure that there had to be a counterexample, so the positive answer surprises me. Now I have to rethink my intuitions on this matter... $\endgroup$
    – Zhen Lin
    May 28, 2015 at 18:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.