In addition to Fernando's answer, note that projective arrows were introduced in 1966 by A. V. Roiter (under the name of projective morphisms)
in the paper *On integral representations belonging to one genus,* Izv. Akad. Nauk SSSR Ser. Mat. 30 (1966), 1315-1324,
see also the English translation: Amer. Math. Soc. Transl. (2) 71 (1968), 49-59.
Roiter defines projective arrows in an abelian category, but actually works with projective arrows in the category of modules over a ring $\Lambda$.
He notices that a morphism of $\Lambda$-modules $p\colon A\to B$ is a projective arrow if and only if $p$ factors via a projective arrow that is an epimorphism.
He uses the notion of a projective arrow in his version of Schanuel's lemma:

**Roiter's lemma.**
Let
$$ 0\to X \to A \to U\to 0,\qquad 0\to Y \to B \to U\to 0 $$
be two short exact sequences, where the morphisms $A\to U$ and $B\to U$ are projective arrows.
Then $B\oplus X \simeq A\oplus Y$.

*Proof:* Let $W$ denote the fibered product of $A$ and $B$ over $U$.
Then $W$ is an extension
$$ 0\to Y\to W\to A\to 0.$$
Since $\varphi\colon A\to U$ is a projective arrow and $\psi\colon B\to U$ is surjective,
$\varphi$ factors as $\psi\circ s$ for some morphism $s\colon A\to B$.
We obtain a morphism $({\rm id}_A,s)\colon A\to W$, which splits the extension $W\to A$.
Thus $W\simeq A\oplus Y$.
Similarly $W\simeq B\oplus X$, hence $B\oplus X \simeq A\oplus Y$, as required.

**Proposition (Roiter).**
Assume that $\Lambda=\mathbb{Z}[\Gamma]$, where $\Gamma$ is a finite group of order $n$.
Let $A$ be a finitely generated free abelian group on which $\Gamma$ acts.
Let $B$ be a finitely generated $\Lambda$-module.
Then for any $\Lambda$-morphism $\varphi\colon A\to B$, the morphism $n\varphi$ is a projective arrow.

*Proof:* Choose an epimorphism $\psi\colon S\twoheadrightarrow B$, where $S$ is a finitely generated free $\Lambda$-module,
then we have an exact sequence
$$ 0\to C\to S\to B\to 0, $$
where $C={\rm ker\,} \psi$, and we obtain the induced exact sequence
$$ 0\to {\rm Hom}(A,C)\to {\rm Hom}(A,S)\to {\rm Hom}(A,B)\overset{\delta}{\longrightarrow}{\rm Ext}_\Lambda^1(A,C).$$
Since $A$ is $\mathbb{Z}$-free,
we have ${\rm Ext}_\Lambda^1(A,C)=H^1(\Gamma, {\rm Hom}_{\mathbb{Z}}(A,C))$.
Since $\#\Gamma=n$, we have $n{\rm Ext}_\Lambda^1(A,C)=nH^1(\Gamma, {\rm Hom}_{\mathbb{Z}}(A,C))=0$, hence $n\delta=0$,
hence $\delta\circ(n\varphi)=(n\delta)\circ\varphi=0$,
and therefore, $n\varphi\in{\rm ker\,}\delta$.
From the exact sequence we see that $n\varphi=\psi\circ x$ for some $x\in {\rm Hom}(A,S)$.
Since $S$ is $\Lambda$-free, the morphism $\psi$ is a projective arrow, hence $n\varphi=\psi\circ x$ is a projective arrow, as required.

**Corollary.**
If $\Lambda$, $\Gamma$, $n$ and $A$ are as above and $U$ is a finite $\Lambda$-module such that $nU=U$,
then *any* $\Lambda$-morphism $A\to U$ is a projective arrow.