Let $X,Y -$compact$X,Y$ be compact metric spaces, $2^X$ the set of all closed subsets of $X$ and $f:X\mapsto Y$$f:X\to Y$ be a 1st class Borel mapping.
Im trying to check Borel class of mapping $G:2^Y\mapsto 2^X$$G:2^Y\to 2^X$. I submit it in a form of 2 compositions: $\theta :2^Y \mapsto X$$\theta :2^Y \to X$ and $\omega:X \mapsto 2^X$$\omega:X \to 2^X$, $G=\theta \circ \omega$. I proofedproved that $\omega$ has 1st Borel class, so if $\theta$ has 1st Borel class, then $G$ too.
I know the following theorem:
Let $X -$compact$X$ be a compact metric space, $Y -$metric$Y$ a metric space. If a multivalued mapping $F:Y\mapsto 2^X$$F:Y\to 2^X$ is upper (lower) semicontinuous, then $F-$$F$ is a 1st class Borel mapping.
But it works with mappings like $F:Y\mapsto 2^X$$F:Y\to 2^X$ and only for upper (lower) semicontinuous mapping.
Q1: AreIs there such a theorem for my example $\theta :2^Y \mapsto X$$\theta :2^Y \to X$?
I have a stupiedstupid idea: Using method of compositions: Let $f^{-1} : Y \mapsto X $$f^{-1} : Y \to X $ and $\sigma : X \mapsto 2^X$$\sigma : X \to 2^X$. By condition $f$ has 1st Borel class. Correct me if I wrong, but $f^{-1}$ has 1st Borel class too. Like $\omega$, $\sigma$ has 1st Borel class. Therefore, we obtain composition of two borelBorel mappings $\theta=f^{-1}\circ\sigma$, so $\theta$ has 1st Borel class.
Q2:Am I mistaken in my reasoning?
