Suppose that $ X, Y $ are Polish spaces, $ A \subset X $ is Borel, $ f : X \to Y $ is Borel, and furthermore, $ f $ is countable-to-one: i.e., $ \forall y \in Y: | f^{-1}(y) | \leq \aleph_0 $. Is it true that $ f(A) $ is Borel?
If $ f $ is injective, this is a corollary of the Lusin-Suslin theorem (see Theorem 15.2 in Kechris).
Yes. This appears as Exercise 18.14 of the same book, and is a consequence of Lusin-Novikov theorem, that states that a projection of a Borel subset of $X\times Y$ with countable sections, is Borel.
The statement of that exercise says that $f(X)$ is Borel, but any Borel $A\subseteq X$ is a standard Borel space with the relative $\sigma$-algebra, and hence admits a Polish topology for which $f\restriction A$ is also Borel, and the result follows.
reference-requestand modified the question. I haven't tried to prove it, as it is the extension of an already deep theorem. – moteutsch May 16 '20 at 13:02