Lemma 10.71.7. Let $R$ be a ring. Let $N$ be an $R$-module. Let $0 \to M' \to M \to M'' \to 0$ be a short exact sequence. Then we get a long exact sequence

**Proof.**
Pick sets of generators $\{ m'_{i'}\} _{i' \in I'}$ and $\{ m''_{i''}\} _{i'' \in I''}$ of $M'$ and $M''$. For each $i'' \in I''$ choose a lift $\tilde m''_{i''} \in M$ of the element $m''_{i''} \in M''$. Set $F' = \bigoplus _{i' \in I'} R$, $F'' = \bigoplus _{i'' \in I''} R$ and $F = F' \oplus F''$. Mapping the generators of these free modules to the corresponding chosen generators gives surjective $R$-module maps $F' \to M'$, $F'' \to M''$, and $F \to M$. We obtain a map of short exact sequences

By the snake lemma we see that the sequence of kernels $0 \to K' \to K \to K'' \to 0$ is short exact sequence of $R$-modules. Hence we can continue this process indefinitely. In other words we obtain a short exact sequence of resolutions fitting into the diagram

Because each of the sequences $0 \to F'_ n \to F_ n \to F''_ n \to 0$ is split exact (by construction) we obtain a short exact sequence of complexes

by applying the $\mathop{\mathrm{Hom}}\nolimits _ R(-, N)$ functor. Thus we get the long exact sequence from the snake lemma applied to this. $\square$

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