Let $X$ be a topological space and $x \in X$. Is the fiber functor $$\mathbf{Top}/X \to \mathbf{Top},\quad (f : Y \to X) \mapsto f^{-1}(x)$$ cocontinuous?

I already checked that coproducts are preserved, but coequalizers are unclear. The composition with the forgetful functor $\mathbf{Top} \to \mathbf{Set}$ is cocontinuous, since the fiber functor $\mathbf{Set}/X \to \mathbf{Set}$ is actually left adjoint to the dependent product along $x : \{\star\} \to X$, which exists in any topos.

I am really interested in $\mathbf{Top}$ here, not a convenient category of spaces.


This is essentially the same as asking whether quotient topologies respect saturated subspaces (i.e., given a space with an equivalence relation, whether the quotient topology on a saturated subspace is the same as the corresponding subspace of the quotient topology for the ambient space), and the answer is no. For a simple example, let $Y=[0,1]\cup[2,3]$, let $A=[2,3)$, let $i:A\to Y$ be the inclusion map and let $c:A\to Y$ be the constant map with value $1$. Then the coequalizer of $i$ and $c$ is the quotient $Z$ of $Y$ that collapses $\{1\}\cup[2,3)$ to a point.

We can further consider this to all be taking place in the category of spaces over $X$ where $X=\{x,y\}$ is a two-point indiscrete space with $\{1\}\cup[2,3)$ mapping to $y$ and $[0,1)\cup\{3\}$ mapping to $x$. If $F$ denotes the functor taking the fiber over $x$, then $F(A)=\emptyset$ and $F(Y)=[0,1)\cup\{3\}$, so the coequalizer of $F(c)$ and $F(i)$ is just $[0,1)\cup\{3\}$ with its usual topology. But $F(Z)$ is $[0,1)\cup\{3\}$ with a different topology, where $3$ is "attached" to the end of $[0,1)$ to form a closed interval (since in the quotient topology of $Z$, every neighborhood of $3$ must contain the equivalence class of $1$).

  • $\begingroup$ Wonderful, thank you! We can also replace $[0,1]$ by $\{1\}$ and obtain a counterexample, right? The coequalizer of $F(c),F(i)$ is the discrete space $\{1,3\}$, but $F$ of the coequalizer of $c,i$ is $\{1,3\}$ with the Sierpinski topology ($\{3\}$ is not open). $\endgroup$ – Martin Brandenburg May 12 at 17:21
  • $\begingroup$ Well, the problem is, to do that, you would need to put $1$ into the fiber over $x$, and then all of $[2,3)$ would be forced to be in the fiber over $x$ as well in order for $c$ to respect the fibers. $\endgroup$ – Eric Wofsey May 12 at 17:24
  • $\begingroup$ Anothre way to say this is that the subspace we look at is required to be saturated with respect to the equivalence relation we're quotienting out, since it needs to be the fiber of a map that descends to the quotient. $\endgroup$ – Eric Wofsey May 12 at 17:26

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