Let $A$ be a commutative ring with unity and consider an $A$-module $M$. Why do we always have the following exact sequence? $$A^{(J)}\rightarrow A^{(I)}\rightarrow M\rightarrow 0$$
(Here $I,J$ are sets and $A^{(I)}$, $A^{(J)}$ are free $A$-modules indexed by $I$ and $J$.)
The map $A^{(I)}\rightarrow M$ is clear: if $\{m_i\}_{i\in I}$ is a set of generators then $(a_i)\mapsto\sum a_im_i$.
But what is the map $A^{(J)}\rightarrow A^{(I)}$?
There's nothing really mysterious. For every module $M$ there exists an epimorphism $g\colon A^{(I)}\to M$, for some set $I$. Consider the kernel $K$ of this epimorphism and apply the same fact, to get a homomorphism $f\colon A^{(J)}\to A^{(I)}$ having $K$ as its image. Then the sequence $$ A^{(J)}\xrightarrow{f}A^{(I)}\xrightarrow{g}M\to 0 $$ is exact.
Just take a set of generators of $K$ (these can be seen as the relations between the generators of $M$).