# Cocontinuity of the fiber functor on topological spaces

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 $$1$$).
• 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). – Martin Brandenburg May 12 at 17:21
• 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. – Eric Wofsey May 12 at 17:24