Section 13.22 (015L): Composition of right derived functors

13.22 Composition of right derived functors

Sometimes we can compute the right derived functor of a composition. Suppose that $\mathcal{A}, \mathcal{B}, \mathcal{C}$ be abelian categories. Let $F : \mathcal{A} \to \mathcal{B}$ and $G : \mathcal{B} \to \mathcal{C}$ be left exact functors. Assume that the right derived functors $RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$, $RG : D^{+}(\mathcal{B}) \to D^{+}(\mathcal{C})$, and $R(G \circ F) : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{C})$ are everywhere defined. Then there exists a canonical transformation

\[ t : R(G \circ F) \longrightarrow RG \circ RF \]

of functors from $D^{+}(\mathcal{A})$ to $D^{+}(\mathcal{C})$, see Lemma 13.14.16. This transformation need not always be an isomorphism.

Lemma 13.22.1. Let $\mathcal{A}, \mathcal{B}, \mathcal{C}$ be abelian categories. Let $F : \mathcal{A} \to \mathcal{B}$ and $G : \mathcal{B} \to \mathcal{C}$ be left exact functors. Assume $\mathcal{A}$, $\mathcal{B}$ have enough injectives. The following are equivalent

$F(I)$ is right acyclic for $G$ for each injective object $I$ of $\mathcal{A}$, and
the canonical map

\[ t : R(G \circ F) \longrightarrow RG \circ RF. \]

is isomorphism of functors from $D^{+}(\mathcal{A})$ to $D^{+}(\mathcal{C})$.

Proof. If (2) holds, then (1) follows by evaluating the isomorphism $t$ on $RF(I) = F(I)$. Conversely, assume (1) holds. Let $A^\bullet $ be a bounded below complex of $\mathcal{A}$. Choose an injective resolution $A^\bullet \to I^\bullet $. The map $t$ is given (see proof of Lemma 13.14.16) by the maps

\[ R(G \circ F)(A^\bullet ) = (G \circ F)(I^\bullet ) = G(F(I^\bullet ))) \to RG(F(I^\bullet )) = RG(RF(A^\bullet )) \]

where the arrow is an isomorphism by Lemma 13.16.7. $\square$

Lemma 13.22.2 (Grothendieck spectral sequence). With assumptions as in Lemma 13.22.1 and assuming the equivalent conditions (1) and (2) hold. Let $X$ be an object of $D^{+}(\mathcal{A})$. There exists a spectral sequence $(E_ r, d_ r)_{r \geq 0}$ consisting of bigraded objects $E_ r$ of $\mathcal{C}$ with $d_ r$ of bidegree $(r, - r + 1)$ and with

\[ E_2^{p, q} = R^ pG(H^ q(RF(X))) \]

Moreover, this spectral sequence is bounded, converges to $H^*(R(G \circ F)(X))$, and induces a finite filtration on each $H^ n(R(G \circ F)(X))$.

For an object $A$ of $\mathcal{A}$ we get $E_2^{p, q} = R^ pG(R^ qF(A))$ converging to $R^{p + q}(G \circ F)(A)$.

Proof. We may represent $X$ by a bounded below complex $A^\bullet $. Choose an injective resolution $A^\bullet \to I^\bullet $. Choose a Cartan-Eilenberg resolution $F(I^\bullet ) \to I^{\bullet , \bullet }$ using Lemma 13.21.2. Apply the second spectral sequence of Lemma 13.21.3. $\square$

Comments (4)

Comment #9980 by Elías Guisado on February 11, 2025 at 05:16

I propose to add the following result to this section. It is the translation of J. Lipman, Corollary 2.2.7, to the Stacks Project definitions and results.

Lemma. Let $\mathcal{A}\xrightarrow{F}\mathcal{B}\xrightarrow{G}\mathcal{C}$ be additive functors between abelian categories. Assume that $RG:D(\mathcal{B})\to D(\mathcal{C})$ is everywhere defined and that every complex $X\in D(\mathcal{A})$ admits a quasi-isomorphism into a complex $A_X$ computing $RF$ . In particular, $RF:D(\mathcal{A})\to D(\mathcal{B})$ is everywhere defined. The following are equivalent: 1. $F(A_X)$ computes $RG$ for every $X\in D(\mathcal{A})$ . 2. $A_X$ computes $R(G\circ F)$ for every $X\in D(\mathcal{A})$ (hence $R(G\circ F):D(\mathcal{A})\to D(\mathcal{C})$ is everywhere defined) and the morphism $t:R(G\circ F)\to RG\circ RF$ coming from Lemma 13.14.16 is an isomorphism.

Proof. Assume 2. Then $R(G\circ F)\to RG\circ RF$ evaluated at $A_X$ becomes $R(G\circ F)(A_X)=G(F(A_X))=G(RF(A_X))\to RG(RF(A_X))$ . Thus $RF(A_X)=F(A_X)$ computes $RG$ . Conversely, suppose $F(A_X)$ computes $RG$ . To show that $A_X$ computes $G\circ F$ , we check the hypotheses of Lemma 13.14.15 for this composite and the class $\mathcal{I}=\{A_X\mid X\in D(\mathcal{A})\}$ . The first hypothesis is clear. To see the second one, given a quasi-isomorphism $A_X\to A_Y$ , $X,Y\in D(\mathcal{A})$ , we have that $F(A_X)\to F(A_Y)$ is a quasi-isomorphism since both $A_X$ and $A_Y$ compute $RF$ (Lemma 13.14.4). Hence $GF(A_X)\to GF(A_Y)$ is a quasi-isomorphism since $F(A_X)$ and $F(A_Y)$ compute $RG$ (Lemma 13.14.4 again). Lastly, $t:R(G\circ F)\to RG\circ RF$ is an isomorphism evaluated at each $A_X$ . Hence it is globally an isomorphism since each $X\in D(\mathcal{A})$ is isomorphic to $A_X$ in $D(\mathcal{A})$ .

Comment #9981 by Elías Guisado on February 11, 2025 at 05:25

(I missed the title in the reference in #9980, it is J. Lipman, Notes on Derived Functors and Grothendieck Duality.)

This might be added as well. It is Görtz, Wedhorn, Algebraic Geometry II, Remark F.177.

Lemma. Let $\mathcal{A}\xrightarrow{F}\mathcal{B}\xrightarrow{G}\mathcal{C}$ be additive functors between abelian categories. Assume that $F$ is exact, $RG:D(\mathcal{B})\to D(\mathcal{C})$ is everywhere defined, and that every complex $X\in D(\mathcal{A})$ admits a quasi-isomorphism into a complex $A_X$ such that $F(A_X)$ computes $RG$ . Then $A_X$ computes $R(G\circ F)$ for every $X\in D(\mathcal{A})$ (hence $R(G\circ F):D(\mathcal{A})\to D(\mathcal{C})$ is everywhere defined) and the morphism $t:R(G\circ F)\to RG\circ RF=RG\circ F$ coming from Lemma 13.14.16 is an isomorphism.

Proof. Just apply #9980. Every complex of $D(\mathcal{A})$ computes $RF$ for $F$ is exact.

Comment #9982 by Elías Guisado on February 11, 2025 at 05:47

(The hypothesis in #9981 of $RG:D(\mathcal{B})\to D(\mathcal{C})$ being everywhere defined may be dropped since it is already implied by the rest.)

Comment #9986 by Elías Guisado on February 12, 2025 at 03:16

Ignore #9982, it's wrong.

The Stacks project

13.22 Composition of right derived functors

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