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Consider the following diagram of left $R$-modules ($R$ is a ring with identity $1_R$) and $R$-linear maps with exact lines:

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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.

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    $\begingroup$ To get the map from $M$ to $M'$ you just need to use that $M$ is projective. $\endgroup$ – Tobias Kildetoft Apr 29 '15 at 10:59
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    $\begingroup$ And to get the map from $L$ to $L'$ afterwards, you need the exactness of the sequences. The map $L\to M \to M' \to N'$ is zero, hence the map $L\to M \to M'$ factors over the kernel of $M' \to N'$, which is $L'$. $\endgroup$ – MooS Apr 29 '15 at 11:13
  • $\begingroup$ @TobiasKildetoft I haven't seen projective modules yet so I guess I can't use it, but thanks anyway. $\endgroup$ – PtF Apr 29 '15 at 11:44
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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.

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