We denote by $A$, $A′$, $B$, $B′$ the four sets.
We are given mapping $u:A′→A$, $v:B→B′$.
We denote by $F(A, B)$ the set maps $A→B$ ，$F(A′, B′)$ the set maps $A′→B′$.
Let $Φ : F(A, B) → F(A′, B′)$ be a mappig which makes $v◦f ◦ u ∈ F(A′, B′)$ correspond with for $f ∈ F(A, B)$.
Prove that we can't prove "$u$ is injective and $v$ is surjective $⇒$ $Φ$ is surjective." without Axiom of choice.
Actually, the $u$ part is misleading, because $f\mapsto f\circ u$ is surjective if $u$ is injective, without the axiom of choice. The interesting part is about $v$ !
So we may as well take $u= id_{A'}$. And now the implication we have is "$v$ surjective implies $f\mapsto v\circ f$ surjective" and we want to see that this implies the axiom of choice.
Consider : $f\mapsto v\circ f$, $F(B', B)\to F(B',B')$.
If you take $id_{B'}\in F(B',B')$, the surjectivity of this map implies that $v$ has a section, i.e. there is a function $B'\to B$ that picks out an antecedent for each element of $B'$.
It's known that "every surjective function has a section" implies the axiom of choice, so there you go