Is every Polish space the quotient of some subset of Baire space?

Question

Question is as in the title:

Is every Polish space the quotient space of some subset of Baire space $\omega^\omega$ ?

Answer 1

Yes. Actually, there are two cases. Take an arbitrary Polish space $X$. If $X=\emptyset$, then the subset of $\omega^\omega$ must be $\emptyset$. If $X$ is nonempty, then it is a quotient (a continuous image) of all of $\omega^\omega$.

You can find a proof in Kechris' Classical Descriptive Set Theory, Sect. 7.E. The main idea is that for $X\neq\emptyset$, you can construct a family $(F_s)_s$ of $F_\sigma$ subsets of $X$ indexed by finite sequences of natural numbers such that

1. $F_\emptyset= X$;
2. $F_s=\bigcup_i F_{s\smallfrown i}=\bigcup_i \overline{F_{s\smallfrown i}}$; and
3. the diameter $F_s$ is less than $2^{-|s|}$,

where $|s|$ is the length of $s$ and $(s_0,\dots,s_k)\smallfrown i = (s_0,\dots,s_k,i)$. For every $y\in \omega^\omega$ such that $\bigcap_m F_{y\restriction m}\neq\emptyset$, this intersection is a singleton and therefore it induces a function between the set $D$ of those $y$s and $X$. This map is continuous, and $D$ is closed. After that, you must use the result that every closed subset of the Baire space is a retract.

Answer 2

Remark: Prof Terraf's answer proves that every Polish space is the continuous surjective image of $w^w$, but it doesn't prove that it is the $quotient$ $space$.