Three Polish spaces $A$, $B$ and $R$ are given, and are equipped with their Borel $σ$-algebras $\mathcal B_A$, $\mathcal B_B$ and $\mathcal B_R$. Let $f\colon A \to B$ and $g\colon A \to R$ be $(\mathcal B_A,\mathcal B_B)$- and $(\mathcal B_A,\mathcal B_R)$-measurable functions. Let's write $p(f,g)$ as a shorthand for the following.
For every Polish space $X$ and all $(\mathcal B_X,\mathcal B_A)$-measurable maps $h,k\colon X\to A$, $$f \circ h = f \circ k \implies g \circ h = g \circ k.$$
My question is, is it true that if $p(f,g)$, then $g$ is $(f^{-1}\mathcal B_B,\mathcal B_R)$-measurable?
Here, $\mathcal B_X$ is the Borel $σ$-algebra of $X$ and $f^{-1}\mathcal B_B = \{f^{-1}(B'):B'\in \mathcal B_B\}$ is the usual pullback $σ$-algebra.
The idea behind this is that $f$ is a sort of extension of a probability space, in the sense that $B$ is extended by $f$ to $A$ (here I don't consider particular measures, though, and $f$ would have additional properties that I think are unnecessary here). $g$ is a $R$-valued random variable, and $p(f,g)$ sort of states that $g$ is not related to the extension and that $g$ is already available in the smaller space $B$. Actually, this is true if the answer to my question is yes, but I'm not even sure where to start to show that (or to confute it).