## Towards the derived category

We want to build a category where given a free resolution \(\cdots F^{-2} \to F^{-1} \to F^0 \to M \to 0\) the module $M$ and the complex $F^\bullet$ are identified.

We write $Ch(\mathcal A)$ for the category of chain complexes of objects of an additive category $\mathcal A$. The objects are chain complexes and the morphisms are chain maps. There are some common variants which impose vanishing conditions:

- $Ch^-(\mathcal A)$ is the category of chain complexes $C$ such there exists an $N$ with $C^i = 0$ for $i > N$. This is the category of
*bounded above*chain complexes. - $Ch^+(\mathcal A)$ is the category of chain complexes $C$ such there exists an $M$ with $C^i = 0$ for $i < M$. This is the category of
*bounded below*chain complexes. - $Ch^b(\mathcal A)$ has objects in the intersection of the previous two. It is the category of
*bounded*chain complexes.

We have a functor \(\begin{aligned} \mathcal A & \to Ch(\mathcal A) \\ A & \mapsto A[0] \end{aligned}\) where \((A[0])^j = \begin{cases} A & j = 0 \\ 0 & j \neq 0 \end{cases}\)

The category of chain complexes also has built in family of auto-equivalences \([i] : Ch(\mathcal A) \to Ch(\mathcal A)\) for $i \in \mathbb{Z}$. The functor $[i]$ is shifts (or translates) a complex $i$ steps to the left \(A[i]^j = A^{i+j}\) and also translates the differential up to sign \(\delta_{A[i]}^j = (-1)^i \delta_A^{i+j}\)

## Quasi-isomorphisms

For a map of chain complexes $\phi: C \to D$, since $\delta_D^i \phi^i = \phi^{i+1} \delta_C^i$, we have well-defined maps \(\phi^i: \ker \delta_C^i \to \ker \delta_D^i \\ \phi^i: \operatorname{im} \delta_C^i \to \operatorname{im} \delta_D^i\) and hence an induced map \(H^i(\phi) : H^i(C) \to H^i(D)\) between the homologies.

**Definition**. A *quasi-isomorphism* is a chain map $\phi : C \to D$ such that \(H^i(\phi) : H^i(C) \to H^i(D)\) is an isomorphism for all $i \in \mathbb{Z}$.

For an example of a quasi-isomorphism, we come back to our free resolution $F \to M$ from above. We can “kink” our complex

and get a map of complexes. Since \(\cdots F^{-2} \to F^{-1} \to F^0 \to M \to 0\) is exact, we know that \(H^i(F) = \begin{cases} M & i = 0 \\ 0 & i \neq 0 \end{cases}\) and moreover the induced map \(H^0(F) \to H^0(M)\) is an isomorphism.

This is the prototypical example of a quasi-isomorphism.

## Derived category

Our goal is the produce a category where quasi-isomorphims are inverted (in a controlled fashion).

Note that this is necessary.

**Example**. The map of complexes

has no inverse since any map $k[x]/(x) \to k[x]$ is necessarily $0$.

**Definition**. The *derived category* of $\mathcal A$, denoted $D(\mathcal A)$, for which there exists a functor \(Q : Ch(\mathcal A) \to D(\mathcal A)\) such that for any quasi-isomorphism $f: C \to D$ $Q(f)$ is an isomorphism and whenever we have a functor $F: Ch(\mathcal A) \to \mathcal D$ which also takes quasi-isomorphisms to isomorphisms there exists a unique $\bar{F}: D(\mathcal A) \to \mathcal D$ and a diagram

which commutes up to natural isomorphism.

There is a general localization result that produces the category, ignoring set-theoretic issues. But the resulting category is difficult to wield.

To give a construction of $D(\mathcal A)$ in some special, but sufficiently universal for us, we will work in two stages.

Perform a well-behaved quotient of $Ch(\mathcal A)$ to obtain a new category $K(\mathcal A)$.

Identify a full subcategory of $\mathcal C \subset K(\mathcal A)$ where:

- every object of $K(\mathcal A)$ is quasi-isomorphic to one in $\mathcal C$ and
- for any object $C$ of $\mathcal C$, the functor $\operatorname{Hom}(C,-)$ (or $\operatorname{Hom}(-,C)$) converts quasi-isomorphisms for isomorphisms.

We start with making $K(\mathcal A)$.

## The homotopy category of chain complexes

The essential complication of managing quasi-isomorphisms is that being a quasi-isomorphism is not a functorial notion. It cannot be expressed via some diagram.

We introduce a related notion that is functorial: homotopy.

**Definition**. A *homotopy* $h$ between two chain maps $f,g: C \to D$ is a collection of maps in $\mathcal A$ \(h^i : C^i \to D^{i-1}\) satisfying \(f^i - g^i = h^{i+1} \delta^i_C + \delta_D^{i-1} h^i\) for all $i \in \mathbb{Z}$. We say $f$ and $g$ are *homotopic*. A morphism is *null-homotopic* if it is homotopic to $0$. A chain complex $C$ is null-homotopic if $1_C$ is null-homotopic.

You can visualize the homotopy identities as saying the sum of thick paths in the diagram

equals $f^i-g^i$.

**Lemma**. Homotopy is an equivalence relation. Moreover, it is a two-sided ideal.

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

For reflexivity, we can take $h^i = 0$. For symmetry, we can replace $h^i$ by $-h^i$. For transitivity, we can add the homotopies. For the ideal condition, if $f$ and $g$ are homotopic via homotopy and $u$ is a chain map then $u \circ h$ is a homotopy between $u \circ f$ and $u \circ g$. Similarly, $h \circ u$ is a homotopy between $f \circ u$ and $g \circ u$. ■

**Lemma**. Homotopic maps induce equal maps on homology.

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

If $$ f^i - g^i = h^{i+1} \delta^i_C + \delta_D^{i-1} h^i $$ and $\delta^i_C c = 0$ then, $$ f^i(c) - g^i(c) = \delta_D^{i-1} h^i(c) $$ meaning that $$ \overline{f^i(c)} = \overline{g^i(c)} \in H^i(D). $$ ■

As a corollary null-homotopic complexes are *acyclic*, $H^i(C) = 0$ for all $i$.

**Definition**. The *homotopy category* $K(\mathcal A)$ has the same objects as $Ch(\mathcal A)$ but the morphism spaces are \(\operatorname{Hom}_{K(\mathcal A)}(C,D) := \operatorname{Hom}_{Ch(\mathcal A)}(C,D)/\text{homotopy}\)

This is well-defined and comes with quotient functor $\pi : Ch(\mathcal A) \to K(\mathcal A)$ which is the identity on the objects and the quotient by the submodule of null-homotopic on morphism spaces.

There is useful enhancement of the morphism spaces we can consider. For two complexes $C$ and $D$, let \(\operatorname{Hom}^l(C,D) := \prod_{i \in \mathbb{Z}} \operatorname{Hom}_{\mathcal A}(C^i,D^{i+l})\) Equip these abelian groups with the structure of chain complex where \((\delta \phi)^j := \delta_D^{j+l} \phi^j - (-1)^l \phi^{j+1} \delta_C^j\) for $\phi \in \operatorname{Hom}^l(C,D)$.

**Lemma**. There are natural isomorphisms \(H^l(\operatorname{Hom}^\bullet(C,D)) \cong \operatorname{Hom}_{K(\mathcal A)}(C,D[l]).\)

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

First note that $$ \operatorname{Hom}^l(C,D) = \operatorname{Hom}^0(C,D) $$ so we can reduce to $l = 0$. For $\phi \in \operatorname{Hom}^0(C,D)$, $\delta \phi = 0$ is the same as $$ \delta_D^{j} \phi^j = \phi^{j+1} \delta_C^j $$ which means $\phi$ is a chain map. For $\phi \in \operatorname{Hom}^{-1}(C,D)$, $$ (\delta \phi)^j = \delta_D^{j-1} \phi^j + \phi^{j+1} \delta_C^j $$ is saying that $\delta \phi$ is null-homotopic with homotopy $\phi$. ■