Characterization of projective modules? 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.
 A: 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 $i>0$.
A: 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.
