## Derived functors

We have a way to derive an additive functor $F : \mathcal A \to \mathcal B$ between abelian categories. If we have enough K-injectives, then the right derived functor is \(\mathbf{R}F(X) := F(I_X)\) where $X \to I_X$ is a quasi-isomorphism with $I_X$ K-injective.

Similarly, if we have enough K-projectives, then the left derived functor is \(\mathbf{L}F(X) := F(P_X)\) where $P_X \to X$ is a quasi-isomorphism with $P_X$ K-projective.

These provide method, at least theoretically, to compute the derived functors:

- Figure out an explicit resolution
- Apply $F$
- Compute homology.

However, as with most mathematical objects, it is also desirable to have an abstract characterization of derived functors.

**Definition**. An additive functor $F: \mathcal T \to \mathcal S$ is called *exact* if there is a natural isomorphism \(F \circ [1] \cong [1] \circ F\) and for any triangle \(X \to Y \to Z \to X[1]\) in $\mathcal T$ we have a triangle \(FX \to FY \to FZ \to FX[1]\) in $\mathcal S$.

Any addivite functor $F: \mathcal A \to \mathcal B$ extends to an exact functor \(F : K(\mathcal A) \to K(\mathcal B)\) since it takes cone complexes to cone complexes.

We would like to descend $F$ to an exact functor between the derived categories but there is only possible if $F$ was already exact as functor between the abelian categories.

Instead, we take the “closest” exact functor to $F$ in a precise way.

Historically, there exists (at least) two specifications for derived functors.

**Definition**. (Verdier) The *right derived functor* of an additive functor $F: \mathcal A \to \mathcal B$ is an exact functor \(\mathbf{R}F: D(\mathcal A) \to D(\mathcal B)\) together with a natural transformation \(j_{\mathcal B} \circ F \to \mathbf{R}F \circ j_{\mathcal B}\) such that for any other exact functor \(G : D(\mathcal A) \to D(\mathcal B)\) and natural transformation \(j_{\mathcal B} \circ F \to G \circ j_{\mathcal B}\) there exists a unique natural transformation $\mathbf{R}F \to G$ making the diagram

There is also a corresponding *left derived functor*. Here \(\begin{aligned} j_{\mathcal A} : \mathcal A & \to D(\mathcal A) \\ A & \mapsto A[0] \end{aligned}\) is the fully-faithful embedding.

Thus, we can view Verdier’s definition of $\mathbf{R}F$ as the “closest” extension of $F$ to a quasi-isomorphism preserving functor.

Another definition, due to Deligne, has the advantage of formulating the existence $\mathbf{R}F(X)$ as a representability problem. Like most representability problems, if we make our category bigger we can solve it.

Fix an object $X$ and note that the class of quasi-isomorphism $X \to X^\prime$ is filtered since we can complete

**Definition**. (Deligne) Let $X$ be an object of $D(\mathcal A)$ then $\mathbf{R}F(X)$ is the object of $D(\mathcal B)$ that represents \(Y \mapsto \operatorname{colim}_{X \underset{q.i}{\to} X^\prime} \operatorname{Hom}_{\mathcal D(\mathcal B)}(Y, FX^\prime )\)

If Deligne’s $\mathbf{R}F$ exists, it satisfies Verdier’s specification also.

The motivation for this definition is the following fact which we won’t prove.

**Proposition**. We have \(\operatorname{Hom}_{D(\mathcal A)}(Y,X) = \operatorname{colim}_{X \underset{q.i}{\to} X^\prime} \operatorname{Hom}_{K(\mathcal A)}(Y,X^\prime)\)

Another fact:

**Propsition**. The full subcategory of objects of $D(\mathcal A)$ where $\mathbf{R}F$ is exists is a triangulated subcategory and is closed under retracts.

Let’s see that under our common assumptions that things agree.

**Proposition**. Assume that $\mathcal A$ has enough injectives and countable products. Then \(\mathbf{R}F(X) \cong F(I_X)\) for a quasi-isomorphism $X \to I_X$ with $I_X$ K-injective.

## **Proof**. (Expand to view)

Given any quasi-isomorphism $X \to X^\prime$, we can find a quasi-isomorphism $X^\prime \to I^\prime$ with $I^\prime$ K-injective. Thus, $$ \operatorname{colim}_{X \underset{q.i}{\to} X^\prime} \operatorname{Hom}(Y, FX^\prime ) \cong \operatorname{colim}_{X \underset{q.i}{\to} I^\prime} \operatorname{Hom}(Y, FI^\prime ) $$ where the later is colimit is take over all quasi-isomorphism $X \to I^\prime$ with $I^\prime$ K-injective. Given two such $X \to I$ and $X \to I^\prime$, we have seen that $I$ and $I^\prime$ are homotopy equivalent. Thus, $FI$ and $FI^\prime$ are homotopy equivalent. In particular, they are quasi-isomorphic. So the colimit is constant and isomorphic to to any $FI$. ■

We get also have the analogous statement for left derived functors.

**Definition**. Given an object $A \in \mathcal A$, the *i-th derived functor* $\mathbf{R}^iF(A)$ is \(\mathbf{R}^iF(A) := H^i(\mathbf{R}F(A))\)

So, as an example, \(\operatorname{Ext}^i_{\mathcal A}(A,B) := H^i(\mathbf{R}\operatorname{Hom}(A,B))\) Given that Hom is a bifunctor, there is possibly some ambiguity in the notation $\mathbf{R}\operatorname{Hom}(A,B)$.

**Lemma**. Assume that $\mathcal A$ has enough projectives and enough injectives. Then, \(\mathbf{R}\operatorname{Hom}(A,B) \cong \operatorname{Hom}(P,B) \cong \operatorname{Hom}(A,I)\) where $P \to A$ is a projective resolution and $B \to I$ is an injective resolution.

## **Proof**. (Expand to view)

We just need to exhibit a quasi-isomorphism. We have

One of the most often used properties of $\mathbf{R}^iF$ or $\mathbf{L}^iF$ is the following.

**Proposition**. Let \(0 \to A \to B \to C \to 0\) be an exact sequence in $\mathcal A$. Then we have a long exact sequence \(\cdots \to \mathbf{R}^iFA \to \mathbf{R}^iFB \to \mathbf{R}^iFC \to \mathbf{R}^{i+1}FA \to \cdots\)

## **Proof**. (Expand to view)

Let $\phi : A \to B$ denote the map. We have a map $C(\phi) \to C$ which is a quasi-isomorphism. Thus, $$ A \to B \to C $$ is isomorphic in $D(\mathcal A)$ to a triangle and hence is one. Since $\mathbf{R}F$ is exact, we have a triangle $$ \mathbf{R}FA \to \mathbf{R}FB \to \mathbf{R}F C \to \mathbf{R}FA[1] $$ Taking the long exact sequence of homology gives the conclusion. ■

We don’t necessarily need to use injectives or projectives to compute derived functors.

**Definition**. An object $A$ of $\mathcal F$ is called *(right) adapted* to $F$ if \(\mathbf{R}^iFA = 0 \ \text{ for } i \neq 0.\)

**Lemma**. Let $A \to A^\prime$ be a quasi-isomorphism of bounded below complexes whose components are adapted to $F$. Then $FA \to FA^\prime$ is a quasi-isomorphism.

## **Proof**. (Expand to view)

[Homework](/homework/02/) ■

Of course any injective object is adapted for any $F$. But, in general, that can be more adapted objects and it may be easier to build resolutions out of them.

**Definition**. Given a left $R$-module $M$ and a right $R$-module $N$, the *i-th Tor modules* is \(\operatorname{Tor}^R_i(M,N) := H^{-i}(M \overset{\mathbf{L}}{\otimes} N)\)

Note that since $\otimes_R$ is right exact we have \(\operatorname{Tor}_0^R(M,N) \cong M \otimes_R N.\)

Similar to the above, we have that Tor is symmetric in the two variables.

**Lemma**. If $P \to M$ and $Q \to N$ are projective resolutions, then $P \otimes_R N$ and $M \otimes_R Q$ are quasi-isomorphic.

## **Proof**. (Expand to view)

We have two maps $$ P \otimes_R N \leftarrow P \otimes_R Q \to M \otimes_R Q $$ which we want to check are quasi-isomorphisms. Here $P \otimes_R Q$ is a complex with $$ (P \otimes_R Q)^i = \bigoplus_{a+b=i} P^a \otimes_R Q^b $$ and $$ \delta_{P \otimes_R Q}(p \otimes q) = \delta_P(p) \otimes q + (-1)^a p \otimes \delta_Q(q) $$ where $p \otimes q \in P^a \otimes_R Q^b$. Consider the subcategory of chain complexes satisfying the condition that $$ X \otimes_R Q \to X \otimes_R N $$ is a quasi-isomorphism. This contains any projective module and is closed under shifts. It is also closed under taking cones. This tells us that any bounded complex of projectives lives in this category. Now let $X$ be a bounded above complex of projectives. Then, to consider any particular homology map, we only need to use $\tau_{\geq j} X$ for some $j$. Thus, $X$ can also be a bounded above complex of projectives and we see the two maps are isomorphisms. ■

Recall that $F$ is flat if $F \otimes_R -$ is exact.

**Proposition**. The following are equivalent:

- $F$ is flat.
- $\operatorname{Tor}_i^R(F,-) = 0$ for $i > 0$.
- $\operatorname{Tor}_1^R(F,-) = 0$.

## **Proof**. (Expand to view)

Assume that $F$ is flat. And take $P \twoheadrightarrow M$ with $P$ projective. Then, we have a short exact sequence $$ 0 \to \operatorname{Tor}_1^R(F,M) \to F \otimes_R K \to F \otimes P \to F \otimes M \to 0 $$ Since $F$ is flat, we know that $$ 0 \to F \otimes_R K \to F \otimes P \to F \otimes M \to 0 $$ is exact also. Thus $\operatorname{Tor}_1^R(F,M) = 0$. We also have $$ \operatorname{Tor}_2^R(F,M) \cong \operatorname{Tor}_1^R(F,K) $$ As we have $\operatorname{Tor}_1$ vanishing for any $M$, we get $\operatorname{Tor}_2^R(F,M) = 0$ for any $M$ also. Moving up the chain we get $\operatorname{Tor}_i^R(F,-) = 0$. If $\operatorname{Tor}_i^R(F,-) = 0$ for $i > 0$, then certainly $\operatorname{Tor}_1^R(F,-) = 0$. Assume that $\operatorname{Tor}_1^R(F,-) = 0$. Take any exact sequence $$ 0 \to A \to B \to C \to 0 $$ and we get an exact sequence $$ \cdots \to \operatorname{Tor}_1^R(F,C) \to F \otimes_R C \to F \otimes_R B \to F \otimes_A \to 0 $$ Since $\operatorname{Tor}_1^R(F,C) = 0$, we have that $$ 0 \to F \otimes_R C \to F \otimes_R B \to F \otimes_R A \to 0 $$ is exact. ■

Let’s compute an example.

**Example**. We really only have one projective resolution so far: the Koszul resolution for $k[x_1,\ldots,x_n]/(x_1,\ldots,x_n)$. In the tensor product, \(K(x_1,\ldots,x_n) \otimes_{k[x_1,\ldots,x_n]} k\) and the maps reduce to $0$ since their images land in the submodule generated by $(x_1,\ldots,x_n)$. So \(H^{-i}\left(K(x_1,\ldots,x_n) \otimes_{k[x_1,\ldots,x_n]} k \right) = K^i \otimes k = k^{n \choose i}\)