The Stacks project

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.

Proposition 13.17.8. Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor of abelian categories.

  1. 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 (13.15.9.1). Moreover, any bounded below complex $A^\bullet $ whose terms are acyclic for $RF$ computes $RF$.

  2. 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 (13.15.9.1). 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 13.16.4 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 13.15.15. 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 right derived functor $RF$ is defined at $J^ n$ and $R^ iF(J^ n) = 0$ for $i \not= 0$. Then $H_ k$ holds trivially for $k \leq n_0$. If $H_ n$ holds, then, using Proposition 13.15.8, we see that $RF$ is defined at $J^{n + 1}$ and $(RF(J^ n), RF(I^{n + 1}), RF(J^{n + 1}))$ is a distinguished triangle of $D^+(\mathcal{B})$. Thus the long exact cohomology sequence (13.11.1.1) associated to this triangle gives an exact sequence

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

and gives that $R^ iF(J^{n + 1}) = 0$ for $i \not\in \{ -1, 0\} $. By Lemma 13.17.1 we see that $R^{-1}F(J^{n + 1}) = 0$. This proves that $H_{n + 1}$ is true hence $H_ k$ holds for all $k$. We also conclude that

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

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

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


Comments (6)

Comment #1274 by JuanPablo on

Hello

In the second paragraph in this proof it says that the short exact sequence produces a distinguished triangle . If I understand this right the distinguished triangle comes from the canonical delta functor, so it is distinguished in and the reason is that the mapping from the mapping cone of into is a quasi-isomorphism.

The problem I have is that the lemma 13.15.12 (tag 05SZ) applies to distinguised triangles in which in this case , so it does not seem to apply here.

Comment #1275 by JuanPablo on

In the second paragraph of this proof, in two places in the enunciate and in the comment above, and should be exchanged. (As ).

Comment #1276 by JuanPablo on

Ok. This problem can be fixed as follows:

is distinguished in so by Lemmas 13.15.4 and 13.15.6 (tags 05SB and 05SC) we get is distinguished in .

Now using the long exact cohomology sequence, obtain by induction in , that is exact and for . So the sequence is exact.

Comment #1301 by on

Yes, this was a kind of subtle mistake. I actually had to move the proposition a bit later because using requires this. I also changed the approach slightly, using Proposition 13.15.8 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

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.


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