How to Modify a Borel function in a Borel way-Self study

I have the following questions: assume that $$X$$ be a standard Borel space (i.e. a Polish space equipped with the $$\sigma$$-algebra generated by open sets) with a (possibly Borel, invariant, ergodic) probability measure $$\mu$$ , $$Y$$ a Borel space. Suppose that $$E$$ be an equivalence relation on $$X$$ (I,'m interested in the case there it is induced by the Borel action of a Polish locslly compact group like, for instance, $$SL_m(\mathbb{Z})$$), that $$X_1\subseteq X$$ has measure $$1$$ and that there exists a Borel function $$f\colon X_1\to Y$$.

$$(1)$$ How can I define (if it exists) a Borel function $$c\colon X\to X$$ in such a way that $$c(X)\subseteq X_1$$, so that $$f\circ c$$ is a Borel function from $$X$$ to $$Y$$ and $$c(x)Ex$$?

$$(2)$$ If the previous question has negative answer, under which assumptions this holds?

My intuition is that $$(a)$$ $$X_1$$ meets every equivalence class and then $$(b)$$ I could define $$c$$ on each such equivalence class to be one fixed element of this nonempty intersection, but I don't know how to prove it (it seems I need the Axiom of Choice to do this; can I avoid it?)

Here is my attempt. About $$(a)$$, I can prove that measure $$1$$ implies dense as follows: assume $$A\subseteq X$$ is a non-empty open set (hence $$\mu (A)>0$$) such that $$A\cap X_1=\emptyset$$. Then $$1=\mu(X_1)<\mu(X_1)+\mu(A)=\mu(X_1\cup A)=1$$, contraddiction. But does it suffice to prove that $$X_1$$ has non-empty intersection with each equivalence class?

What about $$(b)$$?