Let $X$ and $Y$ be compact metrizable spaces with $f:Y\rightarrow X$ an open surjection. Suppose that $G\subseteq Y$ is a closed set. How topologically complicated can the set $\{x\in X : f^{-1}(x)\cap G\text{ is clopen in }f^{-1}(x)\}$ be? Note that it's the same set if we replace clopen with open.


I can find examples where it is closed but not open, open but not closed, and $F\_\sigma$, but I vaguely feel like there should be more complicated examples.



Complexity of set of fibers on which a set is relatively clopen
If a map $f:Y\to X$ is an open surjection, then the inverse map $f^{-1}:X\to \mathcal K(Y)$ to the hyperspace $\mathcal K(Y)$ is continuous. The hyperpsace $\mathcal K(Y)$ is the space of nonpempty compact subsets endowed with the Vietoris topology. 


So, your problem reduces to evaluating the Borel complexity of the set $\mathcal O:=\{K\in\mathcal K(Y): K\cap G$ is open in $K\}$ in $\mathcal K(X)$.


Fix a countable base $\mathcal B$ of the topology of $Y$, which is closed under finite unions. Observe that $\mathcal O=\{K\in\mathcal K(Y):\exists B\in\mathcal B\; B\cap K=G\cap K\}=\bigcup\_{B\in\mathcal B}\mathcal O\_B$ where $\mathcal O\_B:=\{K\in\mathcal K(Y):B\cap K=G\cap K\}$. 


Observe that $\mathcal O\_B=\mathcal O\_B'\cap\mathcal O\_B''$ where $\mathcal O'\_B=\{K\in\mathcal K(Y):B\cap K\subset G\}$ and $\mathcal O''\_B=\{K\in\mathcal K(Y):G\cap K\subset B\}$. 


If $K\notin\mathcal O'\_B$, then $K$ intersects the open set $B\setminus G$ and $\{K'\in\mathcal K(Y):K'\cap(B\setminus G)\ne\emptyset\}$ is an open neighborhood of $K$ in $\mathcal K(Y)\setminus\mathcal O'\_B$, which means that the set $\mathcal O'\_B$ is closed in $\mathcal K(Y)$.


Observe that $K\in\mathcal O''\_B$ is and only if $K\cap G\cap (X\setminus B)=\emptyset$, which means that the set $\mathcal O''\_B$ is open in $\mathcal K(Y)$. So $\mathcal O\_B$ is the intersection of an open and closed sets and $\mathcal O=\bigcup\_{B\in\mathcal B}\mathcal O\_B$ is an $F\_\sigma$-set in $\mathcal K(Y)$ and so is its preimage $\{x\in X:f^{-1}(x)\cap G$ is open in $f^{-1}(x)\}$ in $X$.


