Consider the following diagram of left $R$-modules ($R$ is a ring with identity $1_R$) and $R$-linear maps with exact lines:
where $M$ is free. How do define $R$-lineas maps $M\longrightarrow M^\prime$ and $L\longrightarrow L^\prime$ turning the diagram commutative?
I'm a bit lost as to how the fact $M$ is free will come into play.
Thanks.
If $M$ is not free, the diagram may not be completed as wished, consider the diagram of $\mathbf Z$-modules$\def\Z{\mathbf Z}$ $$ \begin{matrix} 0 & \def\To{\longrightarrow}\To & 0 & \To & \Z/(2) & \To & \Z/(2) & \To & 0 \\ && && && \downarrow {\rm id}&& \\ 0 & \To & \Z & \stackrel{\cdot 2}\To & \Z & \To & \Z/(2) & \To & 0 \end{matrix} $$ The only homomorphism $\Z/(2) \to \Z$ does not make the diagram commutative. But if $M$ is free, you can start by defining $M \to M'$ on a free generating set.