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Suppose $P$ is a left $M$-module, and suppose that for every injective module $E$, there is a $g:P \to E$ making the diagram commute: $$\require{AMScd} \begin{CD} {} @. P \\ @. @VVV \\ E @>\pi>> C @>>> 0 \end{CD} $$

Show that $P$ is projective.

Through some liberal hints, I can see the following:

First assume now we have the exact sequence $A \to B \to 0$ with the following commutative: $$ \begin{CD} {} @. P \\ @. @VVV \\ A @>f>> B @>>> 0 \end{CD} $$ Then we seek a $g':P \to A$ making the diagram commute. Now imbed $A$ in an injective module $E$ (since it is always possible to do this). Since $f$ is surjective $B=A/\!ker(f)$. There is also a well defined map from $E \to E / \!\ker(f)$. Since $A \subset E$, I conclude there is also the inclusion map $i:B \hookrightarrow E / \!\ker(f)$.

So we have someting like this: $$\begin{CD} {} @. P \\ @. @VVV \\ A @>f>> B @>>> 0 \\ @VVV @VVV \\ E @>>> E/\!\ker(f) @>>> 0 \end{CD} $$ By assumption, then there is a $g:P \to E$ making the diagram commute. If I then set $g'=g|A$ am I done? Does this all make sense?

Edit: As Arturo points out, that does not make any sense! Instead I need that $\operatorname{im} g \subset A$ and then I am done!

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    @Arturo - thanks for the diagram fix!2011-06-17
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    Is $g|A$ the restriction of $g$ to $A$? If so, it makes no sense, because the domain of $g$ is $P$, not $E$. Rather, you need to show that the *image* of $g$ (which is contained in $E$) is actually contained in $A$.2011-06-17
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    @Arturo, oh of course you are correct. I need to think a bit on that then!2011-06-17
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    Check Homological Algebra of Cartan.2011-09-15

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I think Arturo actually already gives the answer. The image of $g$ (which is contained in E) is actually contained in $A$.

$g(P)+\mathrm{Ker}(f)=\theta(P)+\mathrm{Ker}(f)\subseteq A$. And $g(p)+\mathrm{Ker}(f)=\theta(p)+\mathrm{Ker}(f)$ for any $p\in P$, here $\theta$ denotes the map from $P$ to $B$..

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If you are happy with a little homological algebra, there is the following alternative argument.

Take any module $A$ and form an exact sequence $0\to A\to E\to B\to 0$ with $E$ injective. Applying $\mathrm{Hom}(P,-)$ to the sequence, we conclude that $\mathrm{Ext}^1(P,A)=0$. Since this holds for all modules $A$, it follows that $P$ is projective.