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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.

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 $fy\in V\subset Y$ 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.

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.

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.

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. To avoid assuming $f$ is closed, the corresponding part of the proof uses (amonst other things) the fact any continuous map from a compact space to 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.

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 $fy\in V\subset Y$ such that $fU\subset V$ and also the induced map $U\to V$ is (universally) closed.

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.

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.

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 $fy\in V\subset Y$ 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.

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. To avoid assuming $f$ is closed, the corresponding part of the proof basically localizes to a compact subset on the base and then uses the fact any continuous map from a compact space to 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.

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 $fy\in V\subset Y$ such that $fU\subset V$ and also the induced map $U\to V$ is (universally) closed.

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.

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. To avoid assuming $f$ is closed, the corresponding part of the proof uses (amonst other things) the fact any continuous map from a compact space to 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.

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 $fy\in V\subset Y$ such that $fU\subset V$ and also the induced map $U\to V$ is (universally) closed.

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.