Lemma 15.84.9. Let $I$ be an ideal of a ring $R$. Let $\alpha : K \to K'$ be a morphism of $D(R)$. Assume

$H^ i(K) = H^ i(K') = 0$ for $i \not\in \{ -1, 0\} $

$H^0(\alpha )$ is an isomorphism and $H^{-1}(\alpha )$ is surjective.

If $K$ satisfies the equivalent conditions (1), (2), and (3) of Lemma 15.84.5, then $K'$ does too.

**Proof.**
Set $M = \mathop{\mathrm{Ker}}(H^{-1}(\alpha ))$. Then $\alpha $ fits into a distinguished trangle

\[ M[1] \to K \to K' \to M[2] \]

For any $R$-module $N$ this determines an exact sequence

\[ \mathop{\mathrm{Ext}}\nolimits ^0_ R(M[1], N) \to \mathop{\mathrm{Ext}}\nolimits ^1_ R(K', N) \to \mathop{\mathrm{Ext}}\nolimits ^1_ R(K, N) \]

Since $\mathop{\mathrm{Ext}}\nolimits ^0_ R(M[1], N) = \mathop{\mathrm{Ext}}\nolimits ^{-1}_ R(M, N) = 0$ we see that $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K', N)$ is a submodule of $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K, N)$. Hence if $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K, N)$ is annihilated by $I$ so is $\mathop{\mathrm{Ext}}\nolimits ^1_ R(K', N)$.
$\square$

## Comments (0)