An effective epimorphism in a category is a morphism that is the coequaliser of its kernel pair, and a universally effective epimorphism is a morphism $f : X \to Y$ such that, for every pullback diagram of the form below, $$\require{AMScd} \begin{CD} X' @>>> X \\ @V{f'}VV @VV{f}V \\ Y' @>>> Y \end{CD}$$ the morphism $f' : X' \to Y'$ is an effective epimorphism.

In the case of the category of topological spaces, an effective epimorphism is precisely a quotient map.

Question. What are the universally effective epimorphisms in the category of topological spaces?

Every universally effective epimorphism is an effective epimorphism, but the converse is known to be false. On the other hand, the class of universally effective epimorphisms contains every open surjection and also every proper surjection, so there are lots of examples. And it is not hard to construct a universally effective epimorphism that is neither open nor proper.

What I'm hoping for is some characterisation of the universally effective epimorphisms in the category of topological spaces as "surjective continuous maps that have property $P$", where $P$ is some property that non-surjective maps may have; it would be even better if $P$ is a generalisation of both open and proper. However, I do not think there is any reason to expect such a characterisation.


1 Answer 1


A surjective map $f: X \twoheadrightarrow Y$ is universally effective iff any of the following two equivalent properties hold:

  1. For every point $y \in Y$ and every covering $\mathcal{U}$ of $f^{-1}(\{y\})$ by open subsets, there exist finitely many subsets $U_1, \ldots, U_n \in \mathcal{U}$ such that $f(U_1)\cup\cdots\cup f(U_n)$ is neighbourhood of $y$.
  2. For every point $y \in Y$ and every filter $\mathcal{F}$ in $Y$ converging to $y$ there exists a point $x \in f^{-1}(\{y\})$ such that $f^{-1}\mathcal{F}$ converges to $x$.

This is proved in the paper "On topological quotient maps preserved by pullbacks or products" by Day and Kelly.


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