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I'm having a hard time with a characterization of projective modules:

A $R$-module $P$ is projective if and only if for every epimorphism $f:I\longrightarrow I^{\prime\prime}$ with $I$ injective and for every morphism $u:P\longrightarrow I^{\prime\prime}$ there exists $v:P\longrightarrow I$ such that $fv=u$.

Obs. The implication $(\Rightarrow)$ follows directly from the definition of projective module.

On the other hand we must start with $$L\stackrel{f}{\longrightarrow }M\longrightarrow 0$$ exact and with a morphism $u:P\longrightarrow M$ where $L$ and $M$ are any modules. Since I can solve my problem when $L$ is injective I thought I should use that every module can be seen as a submodule of an injective but I wasn't able to finish the argument.

Thanks.

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Embed $L$ in an injective module $E$ and consider the push-out diagram $$\require{AMScd} \begin{CD} {} @. {} @. 0 @. 0 \\ @. @. @VVV @VVV \\ 0 @>>> K @>>> L @>>> M @>>> 0 \\ @. @| @VVV @VVV \\ 0 @>>> K @>>> E @>>> N @>>> 0 \\ @. @. @VVV @VVV \\ {} @. {} @. E/L @= E/L \\ @. @. @VVV @VVV \\ {} @. {} @. 0 @. 0 \\ \end{CD} $$ If we apply the functor $\def\H#1{\operatorname{Hom}_R(P,#1)}\H{-}$, we get $$ \begin{CD} {} @. {} @. 0 @. 0 \\ @. @. @VVV @VVV \\ 0 @>>> \H{K} @>a>> \H{L} @>b>> \H{M} \\ @. @| @VcVV @VdVV \\ 0 @>>> \H{K} @>e>> \H{E} @>f>> \H{N} @>>> 0 \\ @. @. @VgVV @VhVV \\ {} @. {} @. \H{E/L} @= \H{E/L} \\ @. @. @VVV \\ {} @. {} @. 0 \\ \end{CD} $$ A standard diagram chasing shows that $b$ is surjective.

The middle row and column are exact by the assumption. Let $x\in\H{M}$. Then $d(x)=f(y)$, so $g(y)=hf(y)=hd(x)=0$. Therefore $y=c(z)$ and $db(z)=fc(z)=f(y)=d(x)$. Injectivity of $d$ yields $x=b(z)$. Thus $b$ is surjective, which is what we needed to prove.

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  • $\begingroup$ Very nice! ( +1 ) $\endgroup$ – peter a g Jun 24 '15 at 20:51
  • $\begingroup$ @peterag Yours is nice too. $\endgroup$ – egreg Jun 24 '15 at 20:53
  • $\begingroup$ Thanks. I accepted this answer because it uses only things I have already studied. $\endgroup$ – PtF Jun 25 '15 at 12:31
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We want to show ${\rm Ext}^1( P, N ) = 0$ for all $N$ (as this is equivalent to $P$ being projective). Accordingly, let $$ 0 \to N \to I^0 \mathop\to^{\delta^0} I^1 \mathop\to^{\delta^1} I^2 \to \cdots$$ be the beginning of an injective resolution for $N$. Suppose $f\colon P \to I^1$ is a cocycle, i.e., $\delta^1\circ f = 0$. Since the sequence is exact, $f$ takes values in the image of $\delta^0$. By the assumption, $f$ lifts to a morphism $\bar f \colon P \to I^0$, i.e., such that $\delta^0 \circ \bar f = f$. Hence, the first ext group is zero...

Comment: the argument actually shows that ${\rm Ext}^i(P,N)=0$ for all $i>0$.

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