Free module and exact sequences

Question

Given $E$ an $A$-module why we can choose a short exact sequence $0\to K\to L\to E\to 0$ s.t $L$ is free?

Can anyone help me? I'm trying to prove this.

Answer 1

You are basically using the universal property of a free module.

Take any generating set $S$ of $E$. Consider the free $A$-module $L = A^{(S)}$ with basis $S$. The inclusion $S \hookrightarrow E$, induces via the universal property of the free module, a unique $A$-linear map $L \rightarrow E$. Since the image of this map contains the generating set $S$ of $E$, hence the map is surjective.

Thus you have an exact sequence $L \rightarrow E \rightarrow 0$. Extend this sequence on the left by taking the kernel of the surjective map, so that you have :

$0 \rightarrow K \rightarrow L \rightarrow E \rightarrow 0$

The universal property of free modules implies in particular that every module is a quotient object of a free module.