\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

The Stacks project

13.17 Higher derived functors

The following simple lemma shows that right derived functors “move to the right”.

Lemma 13.17.1. Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor between abelian categories. Let $K^\bullet \in K^{+}(\mathcal{A})$ and $a \in \mathbf{Z}$.

  1. If $H^ i(K^\bullet ) = 0$ for all $i < a$ and $RF$ is defined at $K^\bullet $, then $H^ i(RF(K^\bullet )) = 0$ for all $i < a$.

  2. If $RF$ is defined at $K^\bullet $ and $\tau _{\leq a}K^\bullet $, then $H^ i(RF(\tau _{\leq a}K^\bullet )) = H^ i(RF(K^\bullet ))$ for all $i \leq a$.

Proof. Assume $K^\bullet $ satisfies the assumptions of (1). Let $K^\bullet \to L^\bullet $ be any quasi-isomorphism. Then it is also true that $K^\bullet \to \tau _{\geq a}L^\bullet $ is a quasi-isomorphism by our assumption on $K^\bullet $. Hence in the category $K^\bullet /\text{Qis}^{+}(\mathcal{A})$ the quasi-isomorphisms $s : K^\bullet \to L^\bullet $ with $L^ n = 0$ for $n < a$ are cofinal. Thus $RF$ is the value of the essentially constant ind-object $F(L^\bullet )$ for these $s$ it follows that $H^ i(RF(K^\bullet )) = 0$ for $i < 0$.

To prove (2) we use the distinguished triangle

\[ \tau _{\leq a}K^\bullet \to K^\bullet \to \tau _{\geq a + 1}K^\bullet \to (\tau _{\leq a}K^\bullet )[1] \]

of Remark 13.12.4 to conclude via Lemma 13.15.6 that $RF$ is defined at $\tau _{\geq a + 1}K^\bullet $ as well and that we have a distinguished triangle

\[ RF(\tau _{\leq a}K^\bullet ) \to RF(K^\bullet ) \to RF(\tau _{\geq a + 1}K^\bullet ) \to RF(\tau _{\leq a}K^\bullet )[1] \]

in $D(\mathcal{B})$. By part (1) we see that $RF(\tau _{\geq a + 1}K^\bullet )$ has vanishing cohomology in degrees $< a + 1$. The long exact cohomology sequence of this distinguished triangle then shows what we want. $\square$

Definition 13.17.2. Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor between abelian categories. Assume $RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$ is everywhere defined. Let $i \in \mathbf{Z}$. The $i$th right derived functor $R^ iF$ of $F$ is the functor

\[ R^ iF = H^ i \circ RF : \mathcal{A} \longrightarrow \mathcal{B} \]

The following lemma shows that it really does not make a lot of sense to take the right derived functor unless the functor is left exact.

Lemma 13.17.3. Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor between abelian categories and assume $RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$ is everywhere defined.

  1. We have $R^ iF = 0$ for $i < 0$,

  2. $R^0F$ is left exact,

  3. the map $F \to R^0F$ is an isomorphism if and only if $F$ is left exact.

Proof. Let $A$ be an object of $\mathcal{A}$. Let $A[0] \to K^\bullet $ be any quasi-isomorphism. Then it is also true that $A[0] \to \tau _{\geq 0}K^\bullet $ is a quasi-isomorphism. Hence in the category $A[0]/\text{Qis}^{+}(\mathcal{A})$ the quasi-isomorphisms $s : A[0] \to K^\bullet $ with $K^ n = 0$ for $n < 0$ are cofinal. Thus it is clear that $H^ i(RF(A[0])) = 0$ for $i < 0$. Moreover, for such an $s$ the sequence

\[ 0 \to A \to K^0 \to K^1 \]

is exact. Hence if $F$ is left exact, then $0 \to F(A) \to F(K^0) \to F(K^1)$ is exact as well, and we see that $F(A) \to H^0(F(K^\bullet ))$ is an isomorphism for every $s : A[0] \to K^\bullet $ as above which implies that $H^0(RF(A[0])) = F(A)$.

Let $0 \to A \to B \to C \to 0$ be a short exact sequence of $\mathcal{A}$. By Lemma 13.12.1 we obtain a distinguished triangle $(A[0], B[0], C[0], a, b, c)$ in $K^{+}(\mathcal{A})$. From the long exact cohomology sequence (and the vanishing for $i < 0$ proved above) we deduce that $0 \to R^0F(A) \to R^0F(B) \to R^0F(C)$ is exact. Hence $R^0F$ is left exact. Of course this also proves that if $F \to R^0F$ is an isomorphism, then $F$ is left exact. $\square$

Lemma 13.17.4. Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor between abelian categories and assume $RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$ is everywhere defined. Let $A$ be an object of $\mathcal{A}$.

  1. $A$ is right acyclic for $F$ if and only if $F(A) \to R^0F(A)$ is an isomorphism and $R^ iF(A) = 0$ for all $i > 0$,

  2. if $F$ is left exact, then $A$ is right acyclic for $F$ if and only if $R^ iF(A) = 0$ for all $i > 0$.

Proof. If $A$ is right acyclic for $F$, then $RF(A[0]) = F(A)[0]$ and in particular $F(A) \to R^0F(A)$ is an isomorphism and $R^ iF(A) = 0$ for $i \not= 0$. Conversely, if $F(A) \to R^0F(A)$ is an isomorphism and $R^ iF(A) = 0$ for all $i > 0$ then $F(A[0]) \to RF(A[0])$ is a quasi-isomorphism by Lemma 13.17.3 part (1) and hence $A$ is acyclic. If $F$ is left exact then $F = R^0F$, see Lemma 13.17.3. $\square$

Lemma 13.17.5. Let $F : \mathcal{A} \to \mathcal{B}$ be a left exact functor between abelian categories and assume $RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$ is everywhere defined. Let $0 \to A \to B \to C \to 0$ be a short exact sequence of $\mathcal{A}$.

  1. If $A$ and $C$ are right acyclic for $F$ then so is $B$.

  2. If $A$ and $B$ are right acyclic for $F$ then so is $C$.

  3. If $B$ and $C$ are right acyclic for $F$ and $F(B) \to F(C)$ is surjective then $A$ is right acyclic for $F$.

In each of the three cases

\[ 0 \to F(A) \to F(B) \to F(C) \to 0 \]

is a short exact sequence of $\mathcal{B}$.

Proof. By Lemma 13.12.1 we obtain a distinguished triangle $(A[0], B[0], C[0], a, b, c)$ in $K^{+}(\mathcal{A})$. As $RF$ is an exact functor and since $R^ iF = 0$ for $i < 0$ and $R^0F = F$ (Lemma 13.17.3) we obtain an exact cohomology sequence

\[ 0 \to F(A) \to F(B) \to F(C) \to R^1F(A) \to \ldots \]

in the abelian category $\mathcal{B}$. Thus the lemma follows from the characterization of acyclic objects in Lemma 13.17.4. $\square$

Lemma 13.17.6. Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor between abelian categories and assume $RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$ is everywhere defined.

  1. The functors $R^ iF$, $i \geq 0$ come equipped with a canonical structure of a $\delta $-functor from $\mathcal{A} \to \mathcal{B}$, see Homology, Definition 12.11.1.

  2. If every object of $\mathcal{A}$ is a subobject of a right acyclic object for $F$, then $\{ R^ iF, \delta \} _{i \geq 0}$ is a universal $\delta $-functor, see Homology, Definition 12.11.3.

Proof. The functor $\mathcal{A} \to \text{Comp}^{+}(\mathcal{A})$, $A \mapsto A[0]$ is exact. The functor $\text{Comp}^{+}(\mathcal{A}) \to D^{+}(\mathcal{A})$ is a $\delta $-functor, see Lemma 13.12.1. The functor $RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$ is exact. Finally, the functor $H^0 : D^{+}(\mathcal{B}) \to \mathcal{B}$ is a homological functor, see Definition 13.11.3. Hence we get the structure of a $\delta $-functor from Lemma 13.4.21 and Lemma 13.4.20. Part (2) follows from Homology, Lemma 12.11.4 and the description of acyclics in Lemma 13.17.4. $\square$

Lemma 13.17.7 (Leray's acyclicity lemma). Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor between abelian categories and assume $RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$ is everywhere defined. Let $A^\bullet $ be a bounded below complex of $F$-acyclic objects. The canonical map

\[ F(A^\bullet ) \longrightarrow RF(A^\bullet ) \]

is an isomorphism in $D^{+}(\mathcal{B})$, i.e., $A^\bullet $ computes $RF$.

Proof. First we claim the lemma holds for a bounded complex of acyclic objects. Namely, it holds for complexes with at most one nonzero object by definition. Suppose that $A^\bullet $ is a complex with $A^ n = 0$ for $n \not\in [a, b]$. Using the “stupid” truncations we obtain a termwise split short exact sequence of complexes

\[ 0 \to \sigma _{\geq a + 1} A^\bullet \to A^\bullet \to \sigma _{\leq a} A^\bullet \to 0 \]

see Homology, Section 12.14. Thus a distinguished triangle $(\sigma _{\geq a + 1} A^\bullet , A^\bullet , \sigma _{\leq a} A^\bullet )$. By induction hypothesis the two outer complexes compute $RF$. Then the middle one does too by Lemma 13.15.12.

Suppose that $A^\bullet $ is a bounded below complex of acyclic objects. To show that $F(A) \to RF(A)$ is an isomorphism in $D^{+}(\mathcal{B})$ it suffices to show that $H^ i(F(A)) \to H^ i(RF(A))$ is an isomorphism for all $i$. Pick $i$. Consider the termwise split short exact sequence of complexes

\[ 0 \to \sigma _{\geq i + 2} A^\bullet \to A^\bullet \to \sigma _{\leq i + 1} A^\bullet \to 0. \]

Note that this induces a termwise split short exact sequence

\[ 0 \to \sigma _{\geq i + 2} F(A^\bullet ) \to F(A^\bullet ) \to \sigma _{\leq i + 1} F(A^\bullet ) \to 0. \]

Hence we get distinguished triangles

\[ \begin{matrix} (\sigma _{\geq i + 2} A^\bullet , A^\bullet , \sigma _{\leq i + 1} A^\bullet ) \\ (\sigma _{\geq i + 2} F(A^\bullet ), F(A^\bullet ), \sigma _{\leq i + 1} F(A^\bullet )) \\ (RF(\sigma _{\geq i + 2} A^\bullet ), RF(A^\bullet ), RF(\sigma _{\leq i + 1} A^\bullet )) \end{matrix} \]

Using the last two we obtain a map of exact sequences

\[ \xymatrix{ H^ i(\sigma _{\geq i + 2} F(A^\bullet )) \ar[r] \ar[d] & H^ i(F(A^\bullet )) \ar[r] \ar[d]^\alpha & H^ i(\sigma _{\leq i + 1} F(A^\bullet )) \ar[r] \ar[d]^\beta & H^{i + 1}(\sigma _{\geq i + 2} F(A^\bullet )) \ar[d] \\ R^ iF(\sigma _{\geq i + 2} A^\bullet ) \ar[r] & R^ iF(A^\bullet ) \ar[r] & R^ iF(\sigma _{\leq i + 1} A^\bullet ) \ar[r] & R^{i + 1}F(\sigma _{\geq i + 2} A^\bullet ) } \]

By the results of the first paragraph the map $\beta $ is an isomorphism. By inspection the objects on the upper left and the upper right are zero. Hence to finish the proof it suffices to show that $R^ iF(\sigma _{\geq i + 2} A^\bullet ) = 0$ and $R^{i + 1}F(\sigma _{\geq i + 2} A^\bullet ) = 0$. This follows immediately from Lemma 13.17.1. $\square$

slogan

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. $\square$

Lemma 13.17.9. Let $F : \mathcal{A} \to \mathcal{B}$ be an exact functor of abelian categories. Then

  1. every object of $\mathcal{A}$ is right acyclic for $F$,

  2. $RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{A})$ is everywhere defined,

  3. $RF : D(\mathcal{A}) \to D(\mathcal{A})$ is everywhere defined,

  4. every complex computes $RF$, in other words, the canonical map $F(K^\bullet ) \to RF(K^\bullet )$ is an isomorphism for all complexes, and

  5. $R^ iF = 0$ for $i \not= 0$.

Proof. This is true because $F$ transforms acyclic complexes into acyclic complexes and quasi-isomorphisms into quasi-isomorphisms. Details omitted. $\square$


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