Let $\mathscr{C}$ be a category and $A, A'\in \mathscr{C}$, $g:A\rightarrow A'$. Suppose that for all $C\in \mathscr{C}$, the map $Mor(C,A)\rightarrow Mor(C, A')$, $u\mapsto g\circ u$ is a bijection. Show that $g$ is an isomorphism.
By setting $C=A'$, I get a morphism $u: A'\rightarrow A$ such that $g\circ u=id_{A'}$. But I don't know how to prove $u\circ g=id_A$.
I appreciate any help.
To conclude you can observe that for every $C \in \mathscr C$ the mapping $$\mathscr C[C,u] \colon \mathscr C[C,A'] \longrightarrow \mathscr C[C,A]$$ is the left inverse to $$\mathscr C[C,g]\colon \mathscr C[C,A] \longrightarrow \mathscr C[C,A']$$ hence it is its inverse: since $\mathscr C[C,g]$ is bijective, hence invertible, its left inverse it's also a right inverse.
In particular this implies $$\mathscr C[A,u \circ g]=\mathscr C[C,u]\circ\mathscr C[C,g]=\text{id}_{\mathscr C[C,A]}$$
is the identity and so $$u\circ g=\mathscr C[A,u \circ g](\text{id}_A)=\text{id}_A$$ which proves your claim.