Proposition 13.16.8 (05TA)—The Stacks project

A functor on Abelian categories is extended to the (bounded below or above) derived category by resolving with a complex that is acyclic for that functor.

Proposition 13.16.8. Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor of abelian categories.

If every object of $\mathcal{A}$ injects into an object acyclic for $RF$, then $RF$ is defined on all of $K^{+}(\mathcal{A})$ and we obtain an exact functor

\[ RF : D^{+}(\mathcal{A}) \longrightarrow D^{+}(\mathcal{B}) \]

see (13.14.9.1). Moreover, any bounded below complex $A^\bullet $ whose terms are acyclic for $RF$ computes $RF$.
If every object of $\mathcal{A}$ is quotient of an object acyclic for $LF$, then $LF$ is defined on all of $K^{-}(\mathcal{A})$ and we obtain an exact functor

\[ LF : D^{-}(\mathcal{A}) \longrightarrow D^{-}(\mathcal{B}) \]

see (13.14.9.1). Moreover, any bounded above complex $A^\bullet $ whose terms are acyclic for $LF$ computes $LF$.

Proof. Assume every object of $\mathcal{A}$ injects into an object acyclic for $RF$. Let $\mathcal{I}$ be the set of objects acyclic for $RF$. Let $K^\bullet $ be a bounded below complex in $\mathcal{A}$. By Lemma 13.15.5 there exists a quasi-isomorphism $\alpha : K^\bullet \to I^\bullet $ with $I^\bullet $ bounded below and $I^ n \in \mathcal{I}$. Hence in order to prove (1) it suffices to show that $F(I^\bullet ) \to F((I')^\bullet )$ is a quasi-isomorphism when $s : I^\bullet \to (I')^\bullet $ is a quasi-isomorphism of bounded below complexes of objects from $\mathcal{I}$, see Lemma 13.14.15. Note that the cone $C(s)^\bullet $ is an acyclic bounded below complex all of whose terms are in $\mathcal{I}$. Hence it suffices to show: given an acyclic bounded below complex $I^\bullet $ all of whose terms are in $\mathcal{I}$ the complex $F(I^\bullet )$ is acyclic.

Say $I^ n = 0$ for $n < n_0$. Setting $J^ n = \mathop{\mathrm{Im}}(d^ n)$ we break $I^\bullet $ into short exact sequences $0 \to J^ n \to I^{n + 1} \to J^{n + 1} \to 0$ for $n \geq n_0$. These sequences induce distinguished triangles $(J^ n, I^{n + 1}, J^{n + 1})$ in $D^+(\mathcal{A})$ by Lemma 13.12.1. For each $k \in \mathbf{Z}$ denote $H_ k$ the assertion: For all $n \leq k$ the object $J^ n$ is in $\mathcal{I}$. Then $H_ k$ holds trivially for $k < n_0$. If $H_ n$ holds, then Lemma 13.14.12 shows that $J^{n + 1}$ is in $\mathcal{I}$ and we have $H_{n + 1}$. By Proposition 13.14.8 we have a distinguished triangle $(RF(J^ n), RF(I^{n + 1}), RF(J^{n + 1}))$. Since $J^ n, I^{n + 1}, J^{n + 1}$ are in $\mathcal{I}$ the long exact cohomology sequence (13.11.1.1) associated to this distinguished triangle collapses to an exact sequence

\[ 0 \to F(J^ n) \to F(I^{n + 1}) \to F(J^{n + 1}) \to 0 \]

This in turn proves that $F(I^\bullet )$ is exact.

The proof in the case of $LF$ is dual. $\square$

Comments (9)

Comment #1274 by JuanPablo on January 27, 2015 at 01:45

Hello

In the second paragraph in this proof it says that the short exact sequence $0\rightarrow \text{Im}(d^n)\rightarrow I^n\rightarrow \text{Im}(d^{n+1})\rightarrow 0$ produces a distinguished triangle $(\text{Im}(d^n), I^n, \text{Im}(d^{n+1}))$ . If I understand this right the distinguished triangle comes from the canonical delta functor, so it is distinguished in $D^+(\mathcal{B})$ and the reason is that the mapping from the mapping cone of $\text{Im}(d^n)[0]\rightarrow I^n[0]$ into $\text{Im}(d^{n+1})[0]$ is a quasi-isomorphism.

The problem I have is that the lemma 13.15.12 (tag 05SZ) applies to distinguised triangles in $\mathcal{D}$ which in this case $\mathcal{D}=K^+(\mathcal{B})$ , so it does not seem to apply here.

Comment #1275 by JuanPablo on January 27, 2015 at 02:17

In the second paragraph of this proof, in two places in the enunciate and in the comment above, $\mathcal{B}$ and $\mathcal{A}$ should be exchanged. (As $F:\mathcal{A}\rightarrow \mathcal{B}$ ).

Comment #1276 by JuanPablo on January 27, 2015 at 15:12

Ok. This problem can be fixed as follows:

$(\text{Im}(d^n)[0],I^n[0],\text{Im}(d^{n+1})[0])$ is distinguished in $D^{+}(\mathcal{A})$ so by Lemmas 13.15.4 and 13.15.6 (tags 05SB and 05SC) we get $(RF(\text{Im}(d^n)[0]),I^n[0],RF(\text{Im}(d^{n+1})[0])$ is distinguished in $D^{+}(\mathcal{B})$ .

Now using the long exact cohomology sequence, obtain by induction in $n$ , that $0\rightarrow R^0F(\text{Im}(d^n))\rightarrow F(I^n)\rightarrow R^0F(\text{Im}(d^{n+1}))\rightarrow 0$ is exact and $R^iF(\text{Im}(d^n))=0$ for $i>0$ . So the sequence $0\rightarrow F(I^0)\rightarrow F(I^1)\rightarrow \dots$ is exact.

Comment #1301 by Johan on February 10, 2015 at 01:29

Yes, this was a kind of subtle mistake. I actually had to move the proposition a bit later because using $R^0F$ requires this. I also changed the approach slightly, using Proposition 13.14.8 instead of a bunch of lemmas at one point, which required me to improve the statement of said proposition. Hopefully it is correct now. Here is the change.

Comment #2601 by Rogier Brussee on June 06, 2017 at 20:55

Suggested slogan: A functor on an Abelian categories is extended to the (bounded below or above) derived category by resolving with a complex that is acyclic for that functor.

Comment #2626 by Johan on July 07, 2017 at 12:36

OK, the slogans were added. See here.

Comment #8404 by Elías Guisado on May 17, 2023 at 10:20

When the proof says " $H_k$ holds trivially for $k\leq n_0$ ", I think it should be $k<n_0$ . For the sake of greater clarity, I propose: (i) in the definition $H_k$ , replacing " $RF$ is defined at $J^n$ " by " $J^n\in\mathcal{I}$ ", (ii) changing "using Proposition 13.14.8, we see that $RF$ is defined at $J^{n + 1}$ " to "using Lemma 13.14.12, we see that $J^{n + 1}\in\mathcal{I}$ ", (iv) substituting the two instances of $R^0F(J^n)$ and $R^0F(J^{n+1})$ by $F(J^n)$ and $F(J^{n+1})$ , respectively.

Comment #8405 by Elías Guisado on May 17, 2023 at 10:32

(Minor typo: in the slogan, "on an Abelian categories" should be "on abelian categories".)

Comment #9018 by Stacks project on April 19, 2024 at 15:27

Thanks, this is indeed better. Fixed here.

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