TITLE: Projective arrows
QUESTION [7 upvotes]: We all know that a projective object in a category $\mathcal{C}$ is an object $P$ in $\mathcal{C}$ such that for every epimorphism $f: X\to Y$ in $\mathcal{C}$ and arrow $g\colon P\to Y$ there is a lift $g'\colon P\to X$ of $f$.
Let us define a projective arrow in $\mathcal{C}$ to be an arrow $g: P\to Y$ such that for every epimorphism $f: X\to Y$ there is a lift $g': P\to X$ of $f$.


What is known about such projective arrows?
What are the projective arrows in the category of groups?
Is this just some kind of limit in disguise?


(I'm using "arrow" instead of "morphism" to avoid confilct with the accepted usage of "projective morphism" in Algebraic Geometry).

REPLY [5 votes]: 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.