The Stacks project

Lemma 13.14.13. Assumptions and notation as in Situation 13.14.1. Let $X, Y$ be objects of $\mathcal{D}$. If $X \oplus Y$ computes $RF$, then $X$ and $Y$ compute $RF$. Similarly for $LF$.

Proof. If $X \oplus Y$ computes $RF$, then $RF(X \oplus Y) = F(X) \oplus F(Y)$. In the proof of Lemma 13.14.7 we have seen that the functor $X/S \times Y/S \to (X \oplus Y)/S$, $(s, s') \mapsto s \oplus s'$ is cofinal. We will use this without further mention. Let $s : X \to X'$ be an element of $S$. Then $F(X) \to F(X')$ has a section, namely,

\[ F(X') \to F(X' \oplus Y) \to RF(X' \oplus Y) = RF(X \oplus Y) = F(X) \oplus F(Y) \to F(X). \]

where we have used Lemma 13.14.4. Hence $F(X') = F(X) \oplus E$ for some object $E$ of $\mathcal{D}'$ such that $E \to F(X' \oplus Y) \to RF(X'\oplus Y) = RF(X \oplus Y)$ is zero (Lemma 13.4.12). Because $RF$ is defined at $X' \oplus Y$ with value $F(X) \oplus F(Y)$ we can find a morphism $t : X' \oplus Y \to Z$ of $S$ such that $F(t)$ annihilates $E$. We may assume $Z = X'' \oplus Y''$ and $t = t' \oplus t''$ with $t', t'' \in S$. Then $F(t')$ annihilates $E$. It follows that $F$ is essentially constant on $X/S$ with value $F(X)$ as desired. $\square$

Comments (1)

Comment #8422 by on

Typo: instead of “ has a section” it should be “has a retraction.” Also, I am having trouble understanding the proof: the first problem I have is with “hence for some object of such that is zero (Lemma 13.4.12).” I do understand why having a retraction implies that it must be isomorphic to an inclusion (this is a property of monomorphisms in pre-triangulated categories), but where does the latter condition on the vanishment of comes from? On the other hand, taking for granted this vanishment condition, I think I understand everything what comes after until “ annihilates ” (this last thing I understand too). But why is it that after “it follows that is essentially constant on with value ”?

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  • 4 comment(s) on Section 13.14: Derived functors in general

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