At the students scientific conference I seen paper where were this propositions:
Let $X$, $Y$ $-$ compact metric spaces, $2^X$ $-$ the set of all closed subsets of $X$.
Proposition 1. Let $f:X→Y$ is single-valued mapping. Consider multivalued mapping $F:2^X→2^Y$, where $F(M)= \overline{f(M)}$, $M∈2^X$. If $f-$ continuous mapping, then $F-$ continuous.
Proposition 2. Let $f:X→Y$ is continuous mapping. Consider multivalued mapping $G:2^Y→2^X$, where $G(N)=\overline{f^{-1} (N)}$, $N∈2^Y$. Then $G:2^Y→2^X$ is upper semicontinuous and 1st class Borel mapping.
Proposition 3. Let $f:X→Y$ is 1st class Borel mapping. Consider multivalued mapping $F:2^X→2^Y$, where $F(M)= \overline{f(M)}$, $M∈2^X$. Then $F-$ 2nd class Borel mapping.
$Question:$ If $f:X→Y$ is 1st class Borel mapping and $G:2^Y→2^X$, where $G(N)=\overline{f^{-1} (N)}$, $N∈2^Y$, which Borel class'll have $G$?
I know counterexample, the Dirichlet function:
Let $X=[0,1], Y=[0,1]$ and $f(x) = \begin{cases}1, & x\in [0,1]\cap \mathbb Q \\0, & x\not \in [0,1]\cap\mathbb Q \end{cases}$
If $f$ - 2nd class Borel mapping then $F:2^X→2^Y$ and $G:2^Y→2^X$, where $F(M)= \overline{f(M)}$, $M∈2^X$ and $G(N)=\overline{f^{-1} (N)}$, $N∈2^Y$, isn't Borel class mappings.
In my opinion, $G$ from my question isn't 2nd Borel class, coz of Dirichlet function. Also, we consider 1st Borel mapping, so $G$ shouldn't have 0 Borel class, thus $G$ has 1st Borel class. But it's only my intuition.
