Lemma 10.71.6. Let $R$ be a ring. Let $M$ be an $R$-module. Let $0 \to N' \to N \to N'' \to 0$ be a short exact sequence. Then we get a long exact sequence

$\begin{matrix} 0 \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N') \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N'') \\ \phantom{0\ } \to \mathop{\mathrm{Ext}}\nolimits ^1_ R(M, N') \to \mathop{\mathrm{Ext}}\nolimits ^1_ R(M, N) \to \mathop{\mathrm{Ext}}\nolimits ^1_ R(M, N'') \to \ldots \end{matrix}$

Proof. Pick a free resolution $F_{\bullet } \to M$. Since each of the $F_ i$ are free we see that we get a short exact sequence of complexes

$0 \to \mathop{\mathrm{Hom}}\nolimits _ R(F_{\bullet }, N') \to \mathop{\mathrm{Hom}}\nolimits _ R(F_{\bullet }, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(F_{\bullet }, N'') \to 0$

Thus we get the long exact sequence from the snake lemma applied to this. $\square$

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