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!