Preimage measurable w.r.t. Effros Borel structure

Question

For a Polish space $X$ let $F(X)$ the set of all closed subsets of $X$. The space $F(X)$ can be equipped with the $\sigma$-algebra generated by the sets of type $\{F \in F(X) : F \cap U \neq \emptyset \}$, where $U$ runs over the open subsets of $X$. This is called the Effros Borel structure and makes $F(X)$ to a standard Borel space (see e.g. Kechris, classical descriptive set theory, chapter 12).

For $X,Y$ Polish and $f : X \to Y$ continuous, the mapping $F(Y) \to F(X) : F \mapsto f^{-1}(F)$ is welldefined.

My question: Is this mapping Borel? (If not: at least $\sigma( \Sigma_1^1)$-measureable?)

What I know so far:

1. In general $F(Y) \to F(X) : F \mapsto f^{-1}(F)$ is not continuous. E.g. $f$ constant and $X=Y=\mathbb{R}$.

2. Kechris states a few examples (proof left as an excerise) of mappings of "similar taste" some of them are Borel some not. E.g. $F \mapsto \overline{f(F)}$ and $(F_1,F_2)\mapsto F_1\cup F_2$ are Borel, whereas $(F_1,F_2)\mapsto F_1\cap F_2$ is not. The main takeaway of this for me is that even "very simple" mappings can fail to be Borel in this context (contrary to functions $\mathbb{R} \to \mathbb{R}$ where you have the rule of thumb that everything written down without using AC will be Borel)

Since my mapping is quite natural (compared to the other stated examples), I suspect that this question is either trivial (and I don't see it) or out of the scope of an exercise.

Has anyone ideas how to tackle that or a reference to a solution?

Answer 1

No, that map is not necessarily Borel.

According to exercise 15.5 in Kechris there is a closed set $C\subseteq\mathcal N^2$ such that the map $x\mapsto C_x$ is not Borel as a map $\mathcal N\to F(\mathcal N)$.

Now consider $\pi\colon\mathcal N^2\to\mathcal N$, the projection on the first coordinate, so that $f=\pi_{|C}\colon C\to\mathcal N$ is a continuous map between Polish spaces, but $f^{-1}(\{x\})=C_x$ and so by the above the map $f^{-1}\colon F(\mathcal N)\to F(C)$ is not Borel either, since it's restriction to $\mathcal N\subseteq F(\mathcal N)$ with the canonical embedding $x\mapsto\{x\}$ is not Borel (and $\mathcal N$ is Borel in $F(\mathcal N)$)