Let $X$ be a nice connected topological space, and $x\in X$ a point. Let $F$ be the functor from covering spaces of $X$ to $\textbf{Sets}$ given by sending $Y\stackrel{f}{\rightarrow} X$ to the fiber $f^{-1}(x)$.
Does this have an adjoint?
(I don't really have any good reason for thinking that it might, but I'm just curious)
If not, is there a good reason it doesn't?
EDIT: What if I were to restrict $F$ to finite covering spaces / finite sets?
Assuming $X$ is nice enough that the usual theory of covering spaces works, $F$ has both a left adjoint and a right adjoint. Note that the fiber of a covering space over $x$ is not just a set but also has an action of the group $G=\pi_1(X,x)$. Moreover, the functor from covering spaces of $X$ to $G$-sets taking a covering space to its fiber over $x$ is an equivalence of categories.
Thinking of covering spaces as just $G$-sets, your functor $F$ is then just the forgetful functor from $G$-sets to sets. The left adjoint is then the functor taking a set $S$ to the free $G$-set on $S$, which can be described explicitly as $G\times S$ with the obvious action. In terms of covering spaces, this takes a set $S$ to $U\times S$ where $U$ is the universal cover.
The forgetful functor also has a right adjoint, namely the functor taking a set $S$ to the set $S^G$ of functions $G\to S$, with $G$ acting by $(g\cdot f)(h)=f(hg)$ for $f\in S^G$, $g,h\in G$. Indeed, if $A$ is a $G$-set, then any set map $f:A\to S$ induces a $G$-map $f_*:A\to S^G$ by $f_*(a)(g)=f(ga)$, and it is straightforward to verify this gives a bijection $\operatorname{Hom}_\mathbf{Sets}(A,S)\to\operatorname{Hom}_{G-\mathbf{Sets}}(A,S^G)$. This right adjoint is not so easy to describe in terms of covering spaces. For instance, if $X=S^1$, then this right adjoint sends a set $S$ with two elements to a covering space whose fiber is $S^\mathbb{Z}$, which is uncountable!