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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).
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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.

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  • $\begingroup$ 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 '15 at 18:18

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