$f: X\rightarrow Y$ s.t. $f^{-1}(\boldsymbol{\Sigma_3^0})\subseteq\boldsymbol{\Sigma_2^0}$

81 Views Asked by At

I'd like to prove or disprove (but I think it's true, maybe it is even trivial but I don't see it) the following statement:

Given two Polish spaces $X,Y$ and a Borel function $f:X\rightarrow Y$ then $f^{-1}(\boldsymbol{\Sigma_3^0}(Y))\subseteq\boldsymbol{\Sigma_2^0}(X)$ iff $f^{-1}(\boldsymbol{\Sigma_2^0}(Y))\subseteq\boldsymbol{\Sigma_2^0}(X)$ and $|\text{ran}(f)| \le \aleph_0$

Now one direction is trivial ($\Leftarrow$), as in fact if the preimages of closed sets are $\boldsymbol{\Delta_2^0}$ and the range is countable then the preimage of any set is $\boldsymbol{\Sigma_2^0}$, but what about the other?

Thanks