The Stacks project

Lemma 13.15.2. In Situation 13.15.1.

  1. Let $X$ be an object of $K^{+}(\mathcal{A})$. The right derived functor of $K(\mathcal{A}) \to D(\mathcal{B})$ is defined at $X$ if and only if the right derived functor of $K^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$ is defined at $X$. Moreover, the values are canonically isomorphic.

  2. Let $X$ be an object of $K^{+}(\mathcal{A})$. Then $X$ computes the right derived functor of $K(\mathcal{A}) \to D(\mathcal{B})$ if and only if $X$ computes the right derived functor of $K^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$.

  3. Let $X$ be an object of $K^{-}(\mathcal{A})$. The left derived functor of $K(\mathcal{A}) \to D(\mathcal{B})$ is defined at $X$ if and only if the left derived functor of $K^{-}(\mathcal{A}) \to D^{-}(\mathcal{B})$ is defined at $X$. Moreover, the values are canonically isomorphic.

  4. Let $X$ be an object of $K^{-}(\mathcal{A})$. Then $X$ computes the left derived functor of $K(\mathcal{A}) \to D(\mathcal{B})$ if and only if $X$ computes the left derived functor of $K^{-}(\mathcal{A}) \to D^{-}(\mathcal{B})$.

Proof. Let $X$ be an object of $K^{+}(\mathcal{A})$. Consider a quasi-isomorphism $s : X \to X'$ in $K(\mathcal{A})$. By Lemma 13.11.5 there exists quasi-isomorphism $X' \to X''$ with $X''$ bounded below. Hence we see that $X/\text{Qis}^+(\mathcal{A})$ is cofinal in $X/\text{Qis}(\mathcal{A})$. Thus it is clear that (1) holds. Part (2) follows directly from part (1). Parts (3) and (4) are dual to parts (1) and (2). $\square$


Comments (2)

Comment #8395 by on

I think that maybe we can give a little bit more detail in "thus it is clear that (1) holds": Consider the commutative diagram: Suppose the left map is essentially constant. Then we still have essentially constancy after postcomposition with the bottom map, Lemma 4.22.8. Hence, since the top map is cofinal, the right map is essentially constant. Conversely, suppose that the right map of the diagram is essentially constant. This means that there is a cocone under with vertex , an object of and a map in satisfying Categories, Definition 4.22.1, (1). Since there is a quasi-isomorphism with (Lemma 13.11.5), by https://stacks.math.columbia.edu/tag/05PT#comment-8394 we can substitute by and assume . We claim that . Since is a (split) monomorphism, we have in , by the proof of Lemma 13.4.12. Hence, since is strictly full and saturated in (13.6.4), we get . Lastly, condition (2) of Categories Definition 4.22.1 for the left functor of last diagram follows from the fact that the top functor is cofinal.

Comment #9005 by on

Well, personally, I would have used Lemma 4.22.11 (and its dual) if I was pressed to explain more. Let's see if we get other comments on this.

There are also:

  • 7 comment(s) on Section 13.15: Derived functors on derived categories

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