Let $X$ be a polish space and $E$ an equivalence relation on $X$ such that $E\subseteq X^2$ is analytic.
Show that for every analytic set $A\subseteq X$ the set $[A]_E=\{x\in X: \exists y\in A: xE\,y\}$ is also analytic.
I struggle with this problem. In general I have to show that $[A]_E$ is a continuous image of a Borel set.
So I have to find a continuous function $f:Y\to X$, where $Y$ is a polish space, and a Borel set $B\subseteq Y$ with $f(B)=[A]_E$.
I know some equivalent statements, but they do not seem to make it easier.
Do you have a hint on how to solve this? Thanks in advance.
Hint:
Since $E$ is analytic, there is a (continuous) map $f : \omega^\omega \to X^2$ whose image is $E$.
Moreover, notice $\pi : X^2 \to X$ which sends $(x,y) \mapsto x$ is continuous.
Then $[X]_E = \{ x \in X ~|~ \exists y . xEy \}$ is nothing more than $\pi[E]$, and so $[X]_E$ is the image of $\pi \circ f : \omega^\omega \to X$ (and thus it is analytic).
Can you modify this argument (or add a step at the end, etc. There's various approaches) to get $[A]_E$? It may be helpful to remember (or prove) that analytic sets are closed under intersections.
I hope this helps ^_^