## The derived category

Definition. The derived category of an abelian category $\mathcal A$ is a category, denoted $D(\mathcal A)$, with a functor $$Q : Ch(\mathcal A) \to D(\mathcal A)$$ which takes quasi-isomorphisms to isomorphisms and such that for any $F : Ch(\mathcal A) \to \mathcal C$ which also inverts quasi-isomorphisms there exists a unique (up to isomorphism) $\overline{F}: D(\mathcal A) \to \mathcal C$ such that

commutes up to natural isomorphism.

There are of course versions $D^\ast(\mathcal A)$ for $\ast \in \lbrace b, -, + \rbrace$.

Suppressing set-theoretic concerns, $D(\mathcal A)$ always exists.

Definition. A class of arrows $\mathcal O$ is called a (left) Ore system if

• $\mathcal O$ is closed under composition
• For $f: X \to Z$ in $\mathcal O$ and $X \to Y$, one can complete the diagram

with $g$ in $\mathcal O$.

• If $\alpha f = 0$ with $f$ in $\mathcal O$, then there is $g$ from $\mathcal O$ with $g \alpha = 0$.

There is also a right Ore system where we reverse arrows.

Theorem. Quasi-isomorphisms form both a left and right Ore system. Consequently, the localization of $Ch(\mathcal A)$ exists.

Haiman has some nice notes from his derived categories course that cover this.

While existence is nice, to get a handle on the derived category, you need to be able to actually compute morphism spaces.

Definition. We let K${}^\ast$-Proj/Inj be the full subcategory of K-Proj/Inj consisting of those complexes with {bounded, bounded above, bounded below} homology for $\ast \in \lbrace b, -, + \rbrace$.

Theorem. Assume that $\mathcal A$ has enough projectives. Then $$D^b(\mathcal A) \cong K^b{\text -}Proj$$ and $$D^-(\mathcal A) \cong K^-{\text -}Proj.$$ If, furthermore, $\mathcal A$ has countable coproducts, then $$D(\mathcal A) \cong K{\text -}Proj.$$

Assume that $\mathcal A$ has enough injectives. Then $$D^b(\mathcal A) \cong K^b{\text -}Inj$$ and $$D^+(\mathcal A) \cong K^-{\text -}Inj.$$ If, furthermore, $\mathcal A$ has countable products, then $$D(\mathcal A) \cong K{\text -}Inj.$$

The first part of constructing $Q$ is to take the quotient map $$\pi : Ch(\mathcal A) \to K(\mathcal A)$$ Then for each complex we choose a K-projective resolution $$r_A : P_A \to A$$ Let $f : A \to B$ be a morphism. Complete $r_B : P_B \to B$ to a triangle
$$P_B \to B \to C \to P_B$$ be it completed to a triangle. Applying $\operatorname{Hom}(P_A,-)$ yields an exact sequence $$\cdots \to \operatorname{Hom}(P_A,C[-1]) \to \operatorname{Hom}(P_A,P_B) \to \operatorname{Hom}(P_A,B) \to \operatorname{Hom}(P_A,C) \to \cdots$$ where \begin{aligned} \operatorname{Hom}(P_A,C[-1]) & = 0 \\ \operatorname{Hom}(P_A,C) & = 0 \end{aligned} Thus, $$r_B \circ - : \operatorname{Hom}(P_A,P_B) \to \operatorname{Hom}(P_A,B)$$ is an isomorphism in the homotopy category. So there exists a unique $$\tilde{f} : P_A \to P_B$$ completing

Given this, we see that \begin{aligned} K(\mathcal A) & \to D(\mathcal A) \\ A & \mapsto P_A \\ f & \mapsto \tilde{f} \end{aligned} is a well-defined functor.

Lemma. If $f$ is a quasi-isomorphism, then $\tilde{f}$ is an isomorphism of chain complexes up to homotopy.

Proof. (Expand to view)

Assume for the moment that $\tilde{f}$ is a quasi-isomorphism. Then we have an exact sequence $$\cdots \to \operatorname{Hom}(P,C[-1]) \to \operatorname{Hom}(P,P_A) \to \operatorname{Hom}(P,P_B) \to \operatorname{Hom}(P,C) \to \cdots$$ where $C$ is the cone over $\tilde{f} : P_A \to P_B$. Since $C$ is acyclic and $P$ is K-projective, the map $$\operatorname{Hom}(P,P_A) \to \operatorname{Hom}(P,P_B)$$ is an isomorphism for all $P$. Thus, so is $\tilde{f} : P_A \to P_B$. Now if $f$ is a quasi-isomorphism and $r_B \tilde{f} = f r_A$ with $r_A,r_B$ also quasi-isomorphisms, then $$H^i(\tilde{f}) = H^i(r_B)^{-1} H^i(f) H^i(r_A)$$ is also an isomorphism for all $i$.

Thus $Q : Ch(\mathcal A) \to D(\mathcal A)$, given by the composition of the two functors, takes quasi-isomorphism to isomorphisms.

Finally, we can check that $Q : Ch(\mathcal A) \to D(\mathcal A)$ satisfies the universal property.

Given $F: Ch(\mathcal A) \to \mathcal C$ that takes quasi-isomorphisms to isomorphisms, it descends to $K(\mathcal A) \to \mathcal C$. We can restrict it to the subcategory K-Proj to get $$\overline{F} : \text{K-Proj} \to \mathcal C$$ And $$\overline{F}Q (A) = F(P_A) \overset{F(r_A)}{\to} F(A)$$ is a natural isomorphism.

Thus, whenever we can find a K-projective resolution $P_A \to A$ for all $A$ in $Ch^\ast(\mathcal A)$, we can use the full subcategory of the homotopy category consisting of those K-projectives quasi-isomorphic to a complex from $Ch^\ast(\mathcal A)$ as a model for the derived category. The case of K-injectives works completely analogously.

## Computing morphisms

As we saw above,

Lemma. Given complexes $A$ and $B$ and K-projective resolutions $P_A \to A$ and $P_B \to B$, then natural map $$\operatorname{Hom}_{K(\mathcal A)}(P_A,P_B) \to \operatorname{Hom}_{K(\mathcal A)}(P_A,B)$$ is an isomorphism. Thus, there is a natural isomorphism $$\operatorname{Hom}_{D(\mathcal A)}(A,B) \cong \operatorname{Hom}_{K(\mathcal A)}(P_A,B)$$

Similarly, given a K-injective resolution $B \to I_B$, we have $$\operatorname{Hom}_{D(\mathcal A)}(A,B) \cong \operatorname{Hom}_{K(\mathcal A)}(A,I_B).$$

The upshot of this observation is that \operatorname{Hom}{K(\mathcal A)}(P_A,B)$$or$$\operatorname{Hom}{K(\mathcal A)}(A,I_B) is much easier to compute than the morphisms in the Ore-localized category.

The main part of the work is finding a resolution that is amenable to computation. While this is definitely still a very serious task in general, it is managable in many examples.

Example. Let $S$ be a commutative ring and $f_1,\ldots,f_c \in S$. Then Koszul complex on $f_1,\ldots,f_c$ is the complex $K(f_1,\ldots,f_c)$ with $$K(f_1,\ldots,f_c)^j := \bigwedge^j S^{\oplus c}$$ and whose differential is given by contraction $\iota_{\phi_f}$ with \begin{aligned} \phi_f : S^{\oplus c} & \to S \\ e_i & \mapsto f_i \end{aligned} for the standard basis $e_i$ of $S^{\oplus c}$.

In general, we get a map $K(f_1,\ldots,f_c) \to S/(f_1,\ldots,f_c)$ but it not a resolution.

Let’s look at a special case where we can check by hand that it is. Take $S = k[x_1,\ldots,x_n]$ with $k$ a field of characteristic zero and $f_1,\ldots,f_n = x_1,\ldots,x_n$.

We can show that $\rho : K(x_1,\ldots,x_n) \to k$ is a quasi-isomorphism. First note that we can split the map $\ell : k \to K(x_1,\ldots,x_n)$ but only as a $k$-module. Then the homotopy $$h = \sum_i \frac{\partial}{\partial x_i} e_i \wedge -$$ shows that $\ell \circ \rho$ is homotopic to the identity.