Lemma 15.84.8. Let $R$ be a ring. 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.
For any $f \in R$ if $f : K \to K$ is $0$, then $f : K' \to K'$ is $0$.
Proof. Set $M = \mathop{\mathrm{Ker}}(H^{-1}(\alpha ))$. Then $\alpha $ fits into a distinguished triangle
Since $K \to K' \xrightarrow {f} K'$ is zero by our assumption, we see that $f : K' \to K'$ factors over a map $M[2] \to K'$. However $\mathop{\mathrm{Hom}}\nolimits (M[2], K') = 0$ for example by Derived Categories, Lemma 13.27.3. $\square$
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