# A simple projective test lemma

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:

$$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{lllll} & & & P & & \\ & & & \da{} & & \\ E & & \ra{\pi} & C & & \ra{} & 0\\ \end{array}$$

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:

$$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{lllll} & & & P & & \\ & & & \da{} & & \\ A & & \ra{f} & B & & \ra{} & 0\\ \end{array}$$

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 $B=A/\operatorname{ker}(f)$. There is also a well defined map from $E \to E / \operatorname{ker}(f)$. Since $A \subset E$, I conclude there is also the inclusion map $i:B \hookrightarrow E / \operatorname{ker}(f)$.

So we have someting like this: $$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{cccccc} & & & P & & \\ & & & \da{} & & \\ A & & \ra{f} & B & & \ra{} & 0\\ \da{} & & & \da{} & & \\ E & & \ra{} & E/ \operatorname{ker} f & & \ra{} & 0\\ \end{array}$$

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! – Juan S Jun 17 '11 at 3:34
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 $A$. – Arturo Magidin Jun 17 '11 at 3:47
@Arturo, oh of course you are correct. I need to think a bit on that then! – Juan S Jun 17 '11 at 3:59
Check Homological Algebra of Cartan. – Frank Murphy Sep 15 '11 at 21:25

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 $B$..