Let $A$ be some countable set, and let $A^{\omega}$ be the set of all sequences $\mathbb N \to A$ equipped with the product topology. A subset $X \subseteq A^{\omega}$ is called Suslin set if there is a countable set $B$, a Borel subset $Y \subseteq B^{\omega}$ and a continuous map $f : B^{\omega} \to A^{\omega}$ such that $X = f(Y)$, i.e. it is the continuous image of some Borel set.
Does anybody has a proof of the following result:
If $X \subseteq A^{\omega}$ is a Suslin set, then there exists a closed set $Y \subseteq \mathbb N^{\omega}$ and a continuous map $f : \mathbb N^{\omega} \to A^{\omega}$ such that $X = f(Y)$.
It is mentioned in a book, but its proof is faulty and I cannot find it anywhere else.
As the countable sets are all equipped with discrete topology, might assume they are all $\omega$. So the problem is reduced to the following: if $Y\subset \omega^\omega$ is Borel, then there exists a continuous function $f$ (in fact continuous bijection) whose domain is a closed subset $F$ of $\omega^\omega$ and $f[F]=Y$ (because then you can compose two maps). You only need to verify the following $$\mathcal{A}=\{Y\subset \omega^\omega: \exists\text{ continuous bijection } f\ \exists\text{ closed }F\subset\omega^\omega \operatorname{dom}(f)=F, f''F=Y\}$$ contains a $\sigma$-algebra containing open sets (i.e. the sub algebra that is closed under taking complements). These sets are called Lusin.
More precisely, try to show the following: