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.

  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 (13.14.9.1). 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 (13.14.9.1). 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

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 13.14.8 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

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 #8404 by on

When the proof says " holds trivially for ", I think it should be . For the sake of greater clarity, I propose: (i) in the definition , replacing " is defined at " by "", (ii) changing "using Proposition 13.14.8, we see that is defined at " to "using Lemma 13.14.12, we see that ", (iv) substituting the two instances of and by and , respectively.

Comment #8405 by on

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


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 05TA. Beware of the difference between the letter 'O' and the digit '0'.