I have some trouble trying to proof the following:
Let $U: \Omega \rightarrow A$ and $V: \Omega \rightarrow B$ be surjective functions. Suppose that we have the condition that for all $\omega, \omega' \in \Omega$, if $V(\omega)=V(\omega')$, then $U(\omega)=U(\omega')$. Show tat there exists a function $f:B\rightarrow A$, such that $f(V)=U$.
What I have seen so far is that in this problem I can use the Axiom of choice of the right inverse but not really sure how to start.
Thanks for your help.
On
Let $f(V(\omega)) := U(\omega)$ for all $\omega \in \Omega$. We first have to check that this defines a function: Suppose that $\omega, \omega' \in \Omega$ are such that $V(\omega) = V(\omega')$. By our assumption this implies $U(\omega) = U(\omega')$ and hence $f(V(\omega)) = f(V(\omega'))$. So $f$ is a function and since $V$ is surjective, the domain of $f$ is all of $B$.
Now $f \circ V = U$, by definition of $f$, and the claim follows.
You want the function $f$ to satisfy, for every $\omega\in\Omega$, $$ f(V(\omega))=U(\omega) $$ There is essentially one way to do it. Let $b\in B$ and choose $\omega_b\in\Omega$ such that $V(\omega_b)=b$. Then define $$ f(b)=U(\omega_b) $$ This doesn't depend on the chosen element, because if $b=V(\omega_b)=V(\omega_b')$, then, by assumption, $U(\omega_b)=U(\omega_b')$.
To be a bit more formal, select a function $g\colon B\to\Omega$ such that $V\circ g$ is the identity on $B$. The existence of this function is a direct consequence of the axiom of choice.
Define $f=U\circ g$ and prove it has the required property.