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 $1$).