Let $R$ be a ring with identity and $P$, $M$ and $N$ three left modules over $R$. Futhermore, suppose $P$ is projetive. Let $f\in \textrm{Hom}_R(M, N)$ and $g\in\textrm{Hom}_R(P, N)$.
How can I show there exists $h\in \textrm{Hom}_R(P, M)$ such that $fh=g$ if and only if $\textrm{Im}(g)\subseteq \textrm{Im}(f)$?
Obs. The implication $(\Rightarrow)$ is immediate. For the converse, I had got nowhere.
Thanks.
The converse is proved by restricting the codomains of the maps $f$ and $g$ to $\textrm{Im}(f)$. This way, we get morphisms (which I denote by the same symbols) $$ f\in\textrm{Hom}_R (M,\textrm{Im}(f)) $$ and $$ g\in\textrm{Hom}_R (P,\textrm{Im}(f)). $$ Note that $g$ can be restricted in this way only because $\textrm{Im}(g)$ is contained in $\textrm{Im}(f)$; we view the restriction as the composition of $g:P\to \textrm{Im}(g)$ and of the inclusion $\textrm{Im}(g) \to \textrm{Im}(f)$.
Now, this restriction of $f$ is surjective onto $\textrm{Im}(f)$, so the fact that $P$ is projective implies the existence of $h\in \textrm{Hom}_R(P, M)$ such that $fh=g$.