Lemma 13.14.13 (05T0)—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. Thus by Categories, Lemma 4.22.11 and by characterization (4) of Categories, Lemma 4.22.9 we know that for any object $W$ in $\mathcal{D}'$ the map

\[ \mathop{\mathrm{Hom}}\nolimits _{\mathcal{D}'}(F(X \oplus Y), W) \longrightarrow \mathop{\mathrm{colim}}\nolimits _{s : X \to X', s' : Y \to Y'} \mathop{\mathrm{Hom}}\nolimits _{\mathcal{D}'}(F(X' \oplus Y'), W) \]

is bijective. Since this arrow is clearly compatible with direct sum decompositions on both sides, we conclude that the map

\[ \mathop{\mathrm{Hom}}\nolimits _{\mathcal{D}'}(F(X), W) \longrightarrow \mathop{\mathrm{colim}}\nolimits _{s : X \to X'} \mathop{\mathrm{Hom}}\nolimits _{\mathcal{D}'}(F(X'), W) \]

is bijective (minor detail omitted). Thus by Categories, Lemma 4.22.9 we conclude $RF$ is defined at $X$ with value $F(X)$. Similarly for $Y$. $\square$

Comments (2)

Comment #8422 by Elías Guisado on May 24, 2023 at 04:59

Typo: instead of “ $F(X)\to F(X')$ 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 $F(X') = F(X) \oplus E$ 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).” I do understand why $F(X)\to F(X')$ having a retraction implies that it must be isomorphic to an inclusion $F(X)\to F(X)\oplus E$ (this is a property of monomorphisms in pre-triangulated categories), but where does the latter condition on the vanishment of $E$ comes from? On the other hand, taking for granted this vanishment condition, I think I understand everything what comes after until “ $F(t')\to E$ annihilates $E$ ” (this last thing I understand too). But why is it that after “it follows that $F$ is essentially constant on $X/S$ with value $F(X)$ ”?

Comment #9046 by Stacks project on May 03, 2024 at 14:53

OK, I changed the proof to use the better characterization of essentially constant systems. An alternative would be to add the Karoubian completion of a (pre)additive category and then deduce this lemma from Lemma 13.14.7. Changes are here.

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