## 13.15 Derived functors on derived categories

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

Situation 13.15.1. Here $F : \mathcal{A} \to \mathcal{B}$ is an additive functor between abelian categories. This induces exact functors

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

See Lemma 13.10.6. 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.

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

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

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

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 13.14.1.

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

Lemma 13.15.2. In Situation 13.15.1.

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.

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 $K^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$.

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.

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 13.15.3. In Situation 13.15.1.

The *right derived functors of $F$* are the partial functors $RF$ associated to cases (1) and (2) of Situation 13.15.1.

The *left derived functors of $F$* are the partial functors $LF$ associated to cases (3) and (4) of Situation 13.15.1.

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

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 13.15.4. 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 $a \in \mathbf{Z}$.

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 $n > a$.

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.**
Proof of part (1). Consider the following induction hypothesis $IH_ n$: There are $P^ j \in \mathcal{P}$, $j \geq n$, with $P^ j = 0$ for $j > a$, maps $d^ j : P^ j \to P^{j + 1}$ for $j \geq n$, and surjective maps $\alpha ^ j : P^ j \to K^ j$ for $j \geq n$ such that the diagram

\[ \xymatrix{ & & P^ n \ar[d]^\alpha \ar[r] & P^{n + 1} \ar[d]^\alpha \ar[r] & P^{n + 2} \ar[d]^\alpha \ar[r] & \ldots \\ \ldots \ar[r] & K^{n - 1} \ar[r] & K^ n \ar[r] & K^{n + 1} \ar[r] & K^{n + 2} \ar[r] & \ldots } \]

is commutative, such that $d^{j + 1} \circ d^ j = 0$ for $j \geq n$, such that $\alpha $ induces isomorphisms $H^ j(K^\bullet ) \to \mathop{\mathrm{Ker}}(d^ j)/\mathop{\mathrm{Im}}(d^{j - 1})$ for $j > n$, and such that $\alpha : \mathop{\mathrm{Ker}}(d^ n) \to \mathop{\mathrm{Ker}}(d_ K^ n)$ is surjective. Then we choose a surjection

\[ P^{n - 1} \longrightarrow K^{n - 1} \times _{K^ n} \mathop{\mathrm{Ker}}(d^ n) = K^{n - 1} \times _{\mathop{\mathrm{Ker}}(d_ K^ n)} \mathop{\mathrm{Ker}}(d^ n) \]

with $P^{n - 1}$ in $\mathcal{P}$. This allows us to extend the diagram above to

\[ \xymatrix{ & P^{n - 1} \ar[d]^\alpha \ar[r] & P^ n \ar[d]^\alpha \ar[r] & P^{n + 1} \ar[d]^\alpha \ar[r] & P^{n + 2} \ar[d]^\alpha \ar[r] & \ldots \\ \ldots \ar[r] & K^{n - 1} \ar[r] & K^ n \ar[r] & K^{n + 1} \ar[r] & K^{n + 2} \ar[r] & \ldots } \]

The reader easily checks that $IH_{n - 1}$ holds with this choice.

We finish the proof of (1) as follows. First we note that $IH_ n$ is true for $n = a + 1$ since we can just take $P^ j = 0$ for $j > a$. Hence we see that proceeding by descending induction we produce a complex $P^\bullet $ with $P^ n = 0$ for $n > a$ consisting of objects from $\mathcal{P}$, and a termwise surjective quasi-isomorphism $\alpha : P^\bullet \to K^\bullet $ as desired.

Proof of part (2). The assumption implies that the morphism $\tau _{\leq a}K^\bullet \to K^\bullet $ (Homology, Section 12.15) is a quasi-isomorphism. Apply part (1) to find $P^\bullet \to \tau _{\leq a}K^\bullet $. The composition $P^\bullet \to K^\bullet $ is the desired quasi-isomorphism.
$\square$

Lemma 13.15.5. 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 $a \in \mathbf{Z}$.

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 $n < a$,

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.**
This lemma is dual to Lemma 13.15.4.
$\square$

Lemma 13.15.6. In Situation 13.15.1. Let $\mathcal{I} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ be a subset with the following properties:

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

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 13.15.5. 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 below 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 13.15.7. In Situation 13.15.1. Let $\mathcal{P} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ be a subset with the following properties:

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

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 13.15.6.
$\square$

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