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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 $f$.

  1. What is known about such projective arrows?
  2. What are the projective arrows in the category of groups?
  3. 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).

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up vote 12 down vote accepted

If you assume that your category $\mathcal C$ has enough projectives, e.g. the category of groups, then your projective arrows are the maps which factor through a projective. Let us check this.

Let $f\colon X\rightarrow Y$ be a projective arrow. Since $\mathcal C$ has enough projectives, we can take an epimorphism $g\colon P\twoheadrightarrow Y$ with projective source, hence $f$ factors through $P$. Conversely, if $f\colon X\rightarrow Y$ factors as

$$f\colon X\stackrel{f''}\rightarrow Q\stackrel{f'}\rightarrow Y$$

with $Q$ projective, then $f'$ can be lifted along any epimorphism $g\colon P\twoheadrightarrow Y$, hence $f$ too.

Maps which factor through a projective are widely studied in homological algebra. The stable category $\underline{\mathcal A}$ of an abelian category $\mathcal A$ is the quotient of $\mathcal A$ by the ideal of morphisms factoring through a projective, i.e. by the ideal of projective maps in your terminology. You can read in wikipedia about the stable category of modules over a ring:

if ${\mathcal A}$ is a Frobenius abelian category, i.e. enough projectives and injectives and both classes of objects coincide, then $\underline{\mathcal A}$ is a triangulated category. All algebraic triangulated categories arise in this way, but allowing ${\mathcal A}$ to be an exact category, not just abelian.

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This was very helpful, thanks Fernando! – Mark Grant Nov 24 '12 at 8:45

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.

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