Let $f\colon X\to Y$, with $Y$ at least $T_1$, be continuous and consider the hyperspace $F(X)$ of closed sets of $X$ with the Fell topology, which has as basis the sets $$\langle K;U_1,\ldots,U_n\rangle=\{F\in F(X)\mid F\cap K=\varnothing\land F\cap U_1\neq\varnothing\land\ldots\land F\cap U_n\neq\varnothing\},$$ with $K$ ranging over compact subsets of $X$ and $U_i$ ranging over open subsets of $X$.
Is the map $Y\to F(X)$ defined by $y\mapsto f^{-1}(y)$ continuous? I'm interested in particular in the case in which $X$ and $Y$ are nice spaces, separable metrizable or Polish with $X$ compact, but I'm also curious about what happens in general.
I think that such a map will be semi-continuous, typically. If $f$ is closed we can at least say that $F: y \to f^{-1}[\{y\}]$ is continuous wrt the "miss closed" subbasic open sets, as then for every $O \supseteq f^{-1}[\{y\}]$ we have an $O'$ around $y$ so that $f^{-1}[O'] \subseteq O$ so $F[O'] \subseteq \{F \in F(X)\mid F \cap (X\setminus O) = \emptyset\}$, etc. making $F$ semi-continuous at $y$ (in the more common Vietoris topology). For open $f$ similar things could be said IIRC. But for general $f$, I'm not so sure $F$ will be even semicontinuous.
($X$ compact then Fell topology = Vietoris topology e.g. )