Let $f:X\to Y$ be a continuous map between topological spaces. Consider the quotient map $\pi:X\twoheadrightarrow X/E$ given by decomposing the fibers of $f$ to their connected components.

In Lemma 6.2.21 his book, Engelking proves that when $f:X\to Y$ is closed and has compact fibers, and $X$ is moreover Hausdorff, then $\pi$ is closed. The end of the proof makes crucial use of $f$ being closed.

On page 102 of the book *Dynamic Topology* by Whyburn & Duda, it is proved that if the connected components of the fibers of $f$ are compact, and $X,Y$ are Hausdorff with $X$ also locally compact, then $\pi$ is closed.

Whyburn's argument does *not* assume $f$ is closed and assumes $X$ is merely *locally* compact Hausdorff instead of compact (which would make $f$ closed). Closedness is circumvented by dealing with a closed subset of a compact fiber and using the fact a continuous image of a compact space into a Hausdorff space is closed.

It somehow feels possible to remove the assumptions on $X,Y$ and replace them by assumptions on $f$ *without* assuming $f$ is closed and still have $\pi$ be closed.


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**Definition.** Say a continuous map $f:X\to Y$ is *locally (universally) closed* if for any open neighborhood $x\in W\subset X$, there exist neighborhoods $x\in U\subset W\subset X$ and $fx\in V\subset Y$ such that $fU\subset V$ and also the induced map $U\to V$ is (universally) closed.

**Remark.** The terminal map $X\to \bf 1$ is locally universally closed iff $X$ is locally compact, so this is a relative notion of local compactness (which plays a crucial role in Whyburn's argument).

**Question.** Suppose $f$ is separated and locally universally closed (suppose also compact fibers if necessary). Is $\pi$ closed?

The separation arguments in the proofs of Engelking and Whyburn seem to carry through when merely assuming $f$ is separated, so it really seems the only challenge is to remove the global assumption of $f$ being closed or the global assumption on the source and target spaces.