## 13.16 Higher derived functors

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

Lemma 13.16.1. Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor between abelian categories. Let $K^\bullet$ be a complex of $\mathcal{A}$ 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 $i < a$.

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 13.14.6 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 13.16.2. 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 13.16.3. 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 13.16.4. 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 13.16.3 part (1) and hence $A$ is acyclic. If $F$ is left exact then $F = R^0F$, see Lemma 13.16.3. $\square$

Lemma 13.16.5. 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 13.16.3) 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 13.16.4. $\square$

Lemma 13.16.6. 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 12.12.1.

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 12.12.3.

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 13.4.22 and Lemma 13.4.21. Part (2) follows from Homology, Lemma 12.12.4 and the description of acyclics in Lemma 13.16.4. $\square$

Lemma 13.16.7 (Leray's acyclicity lemma). Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor between abelian categories. Let $A^\bullet$ be a bounded below complex of right $F$-acyclic objects such that $RF$ is defined at $A^\bullet$1. The canonical map

$F(A^\bullet ) \longrightarrow RF(A^\bullet )$

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

Proof. Let $A^\bullet$ be a bounded complex of right $F$-acyclic objects. We claim that $RF$ is defined at $A^\bullet$ and that $F(A^\bullet ) \to RF(A^\bullet )$ is an isomorphism in $D^+(\mathcal{B})$. Namely, it holds for complexes with at most one nonzero right $F$-acyclic object for example by Lemma 13.16.4. Next, suppose that $A^ n = 0$ 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 12.15. Thus a distinguished triangle $(\sigma _{\geq a + 1} A^\bullet , A^\bullet , \sigma _{\leq a} A^\bullet )$. By induction hypothesis $RF$ is defined for the two outer complexes and these complexes compute $RF$. Then the same is true for the middle one by Lemma 13.14.12.

Suppose that $A^\bullet$ is a bounded below complex of acyclic objects such that $RF$ is defined at $A^\bullet$. To show that $F(A^\bullet ) \to RF(A^\bullet )$ is an isomorphism in $D^{+}(\mathcal{B})$ it suffices to show that $H^ i(F(A^\bullet )) \to H^ i(RF(A^\bullet ))$ 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

$(\sigma _{\geq i + 2} A^\bullet , A^\bullet , \sigma _{\leq i + 1} A^\bullet ) \quad \text{and}\quad (\sigma _{\geq i + 2} F(A^\bullet ), F(A^\bullet ), \sigma _{\leq i + 1} F(A^\bullet ))$

Since $RF$ is defined at $A^\bullet$ (by assumption) and at $\sigma _{\leq i + 1}A^\bullet$ (by the first paragraph) we see that $RF$ is defined at $\sigma _{\geq i + 1}A^\bullet$ and we get a distinghuished triangle

$(RF(\sigma _{\geq i + 2} A^\bullet ), RF(A^\bullet ), RF(\sigma _{\leq i + 1} A^\bullet ))$

See Lemma 13.14.6. Using these distinguished triangles we obtain a map of exact sequences

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

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 $H^ i(RF(\sigma _{\geq i + 2} A^\bullet )) = 0$ and $H^{i + 1}(RF(\sigma _{\geq i + 2} A^\bullet )) = 0$. This follows immediately from Lemma 13.16.1. $\square$

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 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 13.14.8, 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 13.16.1 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 13.16.9. 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. $RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$ is everywhere defined,

3. $RF : D(\mathcal{A}) \to D(\mathcal{B})$ 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. $\square$

[1] For example this holds if $RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$ is everywhere defined.

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