
## 13.16 Derived functors on derived categories

In practice derived functors come about most often when given an additive functor between abelian categories.

Situation 13.16.1. Here $F : \mathcal{A} \to \mathcal{B}$ is an additive functor between abelian categories. This induces exact functors

$F : K(\mathcal{A}) \to K(\mathcal{B}), \quad K^{+}(\mathcal{A}) \to K^{+}(\mathcal{B}), \quad K^{-}(\mathcal{A}) \to K^{-}(\mathcal{B}).$

We also denote $F$ the composition $K(\mathcal{A}) \to D(\mathcal{B})$, $K^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$, and $K^{-}(\mathcal{A}) \to D^-(\mathcal{B})$ of $F$ with the localization functor $K(\mathcal{B}) \to D(\mathcal{B})$, etc. This situation leads to four derived functors we will consider in the following.

1. The right derived functor of $F : K(\mathcal{A}) \to D(\mathcal{B})$ relative to the multiplicative system $\text{Qis}(\mathcal{A})$.

2. The right derived functor of $F : K^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$ relative to the multiplicative system $\text{Qis}^{+}(\mathcal{A})$.

3. The left derived functor of $F : K(\mathcal{A}) \to D(\mathcal{B})$ relative to the multiplicative system $\text{Qis}(\mathcal{A})$.

4. The left derived functor of $F : K^{-}(\mathcal{A}) \to D^{-}(\mathcal{B})$ relative to the multiplicative system $\text{Qis}^-(\mathcal{A})$.

Each of these cases is an example of Situation 13.15.1.

Some of the ambiguity that may arise is alleviated by the following.

Lemma 13.16.2. In Situation 13.16.1.

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). $\square$

Given an object $A$ of an abelian category $\mathcal{A}$ we get a complex

$A[0] = ( \ldots \to 0 \to A \to 0 \to \ldots )$

where $A$ is placed in degree zero. Hence a functor $\mathcal{A} \to K(\mathcal{A})$, $A \mapsto A[0]$. Let us temporarily say that a partial functor is one that is defined on a subcategory.

Definition 13.16.3. In Situation 13.16.1.

1. The right derived functors of $F$ are the partial functors $RF$ associated to cases (1) and (2) of Situation 13.16.1.

2. The left derived functors of $F$ are the partial functors $LF$ associated to cases (3) and (4) of Situation 13.16.1.

3. An object $A$ of $\mathcal{A}$ is said to be right acyclic for $F$, or acyclic for $RF$ if $A[0]$ computes $RF$.

4. An object $A$ of $\mathcal{A}$ is said to be left acyclic for $F$, or acyclic for $LF$ if $A[0]$ computes $LF$.

The following few lemmas give some criteria for the existence of enough acyclics.

Lemma 13.16.4. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{I} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ be a subset containing $0$ such that every object of $\mathcal{A}$ is a subobject of an element of $\mathcal{I}$. Let $a \in \mathbf{Z}$.

1. Given $K^\bullet$ with $K^ n = 0$ for $n < a$ there exists a quasi-isomorphism $K^\bullet \to I^\bullet$ with $K^ n \to I^ n$ injective and $I^ n \in \mathcal{I}$ for all $n$ and $I^ n = 0$ for $n < a$,

2. Given $K^\bullet$ with $H^ n(K^\bullet ) = 0$ for $n < a$ there exists a quasi-isomorphism $K^\bullet \to I^\bullet$ with $I^ n \in \mathcal{I}$ and $I^ n = 0$ for $n < a$.

Proof. Proof of part (1). Consider the following induction hypothesis $IH_ n$: There are $I^ j \in \mathcal{I}$, $j \leq n$ almost all zero, maps $d^ j : I^ j \to I^{j + 1}$ for $j < n$ and injective maps $\alpha ^ j : K^ j \to I^ j$ for $j \leq n$ such that the diagram

$\xymatrix{ \ldots \ar[r] & K^{n - 1} \ar[d]^\alpha \ar[r] & K^ n \ar[d]^\alpha \ar[r] & K^{n + 1} \ar[r] & \ldots \\ \ldots \ar[r] & I^{n - 1} \ar[r] & I^ n & & }$

is commutative, such that $d^ j \circ d^{j - 1} = 0$ for $j < n$ and such that $\alpha$ induces isomorphisms $H^ j(K^\bullet ) \to \mathop{\mathrm{Ker}}(d^ j)/\mathop{\mathrm{Im}}(d^{j - 1})$ for $j < n$. Note that this implies

13.16.4.1
$$\label{derived-equation-new-star} \alpha (\mathop{\mathrm{Im}}(d_ K^{n - 1})) \subset \alpha (\mathop{\mathrm{Ker}}(d_ K^ n)) \cap \mathop{\mathrm{Im}}(d^{n - 1}) \subset \alpha (K^ n) \cap \mathop{\mathrm{Im}}(d^{n - 1}).$$

If these inclusions are not equalities, then choose an injection

$I^ n \oplus K^ n/\mathop{\mathrm{Im}}(d_ K^{n - 1}) \longrightarrow I$

with $I \in \mathcal{I}$. Denote $\alpha ' : K^ n \to I$ the map obtained by composing $\alpha \oplus 1 : K^ n \to I^ n \oplus K^ n/\mathop{\mathrm{Im}}(d_ K^{n - 1})$ with the displayed injection. Denote $d' : I^{n - 1} \to I$ the composition $I^{n - 1} \to I^ n \to I$ of $d^{n - 1}$ by the inclusion of the first summand. Then $\alpha '(K^ n) \cap \mathop{\mathrm{Im}}(d') = \alpha '(\mathop{\mathrm{Im}}(d_ K^{n - 1}))$ simply because the intersection of $\alpha '(K^ n)$ with the first summand of $I^ n \oplus K^ n/\mathop{\mathrm{Im}}(d_ K^{n - 1})$ is equal to $\alpha '(\mathop{\mathrm{Im}}(d_ K^{n - 1}))$. Hence, after replacing $I^ n$ by $I$, $\alpha$ by $\alpha '$ and $d^{n - 1}$ by $d'$ we may assume that we have equality in Equation (13.16.4.1). Once this is the case consider the solid diagram

$\xymatrix{ K^ n/\mathop{\mathrm{Ker}}(d_ K^ n) \ar[r] \ar[d] & K^{n + 1} \ar@{..>}[d] \\ I^ n/(\mathop{\mathrm{Im}}(d^{n - 1}) + \alpha (\mathop{\mathrm{Ker}}(d_ K^ n))) \ar@{..>}[r] & M }$

The horizontal arrow is injective by fiat and the vertical arrow is injective as we have equality in (13.16.4.1). Hence the push-out $M$ of this diagram contains both $K^{n + 1}$ and $I^ n/(\mathop{\mathrm{Im}}(d^{n - 1}) + \alpha (\mathop{\mathrm{Ker}}(d_ K^ n)))$ as subobjects. Choose an injection $M \to I^{n + 1}$ with $I^{n + 1} \in \mathcal{I}$. By construction we get $d^ n : I^ n \to I^{n + 1}$ and an injective map $\alpha ^{n + 1} : K^{n + 1} \to I^{n + 1}$. The equality in Equation (13.16.4.1) and the construction of $d^ n$ guarantee that $\alpha : H^ n(K^\bullet ) \to \mathop{\mathrm{Ker}}(d^ n)/\mathop{\mathrm{Im}}(d^{n - 1})$ is an isomorphism. In other words $IH_{n + 1}$ holds.

We finish the proof of by the following observations. First we note that $IH_ n$ is true for $n = a$ since we can just take $I^ j = 0$ for $j < a$ and $K^ a \to I^ a$ an injection of $K^ a$ into an element of $\mathcal{I}$. Next, we note that in the proof of $IH_ n \Rightarrow IH_{n + 1}$ we only modified the object $I^ n$, the map $d^{n - 1}$ and the map $\alpha ^ n$. Hence we see that proceeding by induction we produce a complex $I^\bullet$ with $I^ n = 0$ for $n < a$ consisting of objects from $\mathcal{I}$, and a termwise injective quasi-isomorphism $\alpha : K^\bullet \to I^\bullet$ as desired.

Proof of part (2). The assumption implies that the morphism $K^\bullet \to \tau _{\geq a}K^\bullet$ (Homology, Section 12.14) is a quasi-isomorphism. Apply part (1) to find $\tau _{\geq a}K^\bullet \to I^\bullet$. The composition $K^\bullet \to I^\bullet$ is the desired quasi-isomorphism. $\square$

Lemma 13.16.5. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{P} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ be a subset containing $0$ such that every object of $\mathcal{A}$ is a quotient of an element of $\mathcal{P}$. Let $a \in \mathbf{Z}$.

1. Given $K^\bullet$ with $K^ n = 0$ for $n > a$ there exists a quasi-isomorphism $P^\bullet \to K^\bullet$ with $P^ n \in \mathcal{P}$ and $P^ n \to K^ n$ surjective for all $n$ and $P^ n = 0$ for $n > a$.

2. Given $K^\bullet$ with $H^ n(K^\bullet ) = 0$ for $n > a$ there exists a quasi-isomorphism $P^\bullet \to K^\bullet$ with $P^ n \in \mathcal{P}$ for all $n$ and $P^ n = 0$ for $n > a$.

Proof. This lemma is dual to Lemma 13.16.4. $\square$

Lemma 13.16.6. In Situation 13.16.1. Let $\mathcal{I} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ be a subset with the following properties:

1. every object of $\mathcal{A}$ is a subobject of an element of $\mathcal{I}$,

2. for any short exact sequence $0 \to P \to Q \to R \to 0$ of $\mathcal{A}$ with $P, Q \in \mathcal{I}$, then $R \in \mathcal{I}$, and $0 \to F(P) \to F(Q) \to F(R) \to 0$ is exact.

Then every object of $\mathcal{I}$ is acyclic for $RF$.

Proof. We may add $0$ to $\mathcal{I}$ if necessary. Pick $A \in \mathcal{I}$. Let $A[0] \to K^\bullet$ be a quasi-isomorphism with $K^\bullet$ bounded below. Then we can find a quasi-isomorphism $K^\bullet \to I^\bullet$ with $I^\bullet$ bounded below and each $I^ n \in \mathcal{I}$, see Lemma 13.16.4. Hence we see that these resolutions are cofinal in the category $A[0]/\text{Qis}^{+}(\mathcal{A})$. To finish the proof it therefore suffices to show that for any quasi-isomorphism $A[0] \to I^\bullet$ with $I^\bullet$ bounded above and $I^ n \in \mathcal{I}$ we have $F(A)[0] \to F(I^\bullet )$ is a quasi-isomorphism. To see this suppose that $I^ n = 0$ for $n < n_0$. Of course we may assume that $n_0 < 0$. Starting with $n = n_0$ we prove inductively that $\mathop{\mathrm{Im}}(d^{n - 1}) = \mathop{\mathrm{Ker}}(d^ n)$ and $\mathop{\mathrm{Im}}(d^{-1})$ are elements of $\mathcal{I}$ using property (2) and the exact sequences

$0 \to \mathop{\mathrm{Ker}}(d^ n) \to I^ n \to \mathop{\mathrm{Im}}(d^ n) \to 0.$

Moreover, property (2) also guarantees that the complex

$0 \to F(I^{n_0}) \to F(I^{n_0 + 1}) \to \ldots \to F(I^{-1}) \to F(\mathop{\mathrm{Im}}(d^{-1})) \to 0$

is exact. The exact sequence $0 \to \mathop{\mathrm{Im}}(d^{-1}) \to I^0 \to I^0/\mathop{\mathrm{Im}}(d^{-1}) \to 0$ implies that $I^0/\mathop{\mathrm{Im}}(d^{-1})$ is an element of $\mathcal{I}$. The exact sequence $0 \to A \to I^0/\mathop{\mathrm{Im}}(d^{-1}) \to \mathop{\mathrm{Im}}(d^0) \to 0$ then implies that $\mathop{\mathrm{Im}}(d^0) = \mathop{\mathrm{Ker}}(d^1)$ is an elements of $\mathcal{I}$ and from then on one continues as before to show that $\mathop{\mathrm{Im}}(d^{n - 1}) = \mathop{\mathrm{Ker}}(d^ n)$ is an element of $\mathcal{I}$ for all $n > 0$. Applying $F$ to each of the short exact sequences mentioned above and using (2) we observe that $F(A)[0] \to F(I^\bullet )$ is an isomorphism as desired. $\square$

Lemma 13.16.7. In Situation 13.16.1. Let $\mathcal{P} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ be a subset with the following properties:

1. every object of $\mathcal{A}$ is a quotient of an element of $\mathcal{P}$,

2. for any short exact sequence $0 \to P \to Q \to R \to 0$ of $\mathcal{A}$ with $Q, R \in \mathcal{P}$, then $P \in \mathcal{P}$, and $0 \to F(P) \to F(Q) \to F(R) \to 0$ is exact.

Then every object of $\mathcal{P}$ is acyclic for $LF$.

Proof. Dual to the proof of Lemma 13.16.6. $\square$

Comment #455 by Keenan Kidwell on

In 05T4, there is a superscript minus missing in the target of the third functor of the sentence beginning "We also denote $F$..."

Comment #512 by Keenan Kidwell on

In 05TA, the roles of $\mathcal{B}$ and $\mathcal{A}$ should be switched.

Comment #657 by Fan Zheng on

Why additive functors induce exact functors on the homotopy category of chains?

Comment #661 by on

@#657. This is true because an additive functor $F$ transforms a termwise split short exact sequence of complexes into a termwise split short exact sequence of complexes and an "exact functor between triangulated categories" is an addtive functor which transforms distinguished triangles into distinguished triangles. Anyway, the precise statement is Lemma 13.10.6.

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