# The derived category is additive

Let $$\mathcal C$$ be an abelian category. One way to see the derived category $$D(\mathcal C)$$ is that it has

• the same objects as $$\operatorname{Ch}(\mathcal C)$$,
• roofs $$A\xleftarrow{\simeq}Z_1\rightarrow Z_2\xleftarrow{\simeq}Z_3\rightarrow\cdots \xleftarrow{\simeq}Z_n\rightarrow B$$ as morphisms.

To see that $$D(\mathcal C)$$ is additive, it suffices to show that it contains finite biproducts, for then we can define the addition of morphisms in terms of $$\oplus$$. So the goal is to find a biproduct of two objects $$A, B\in D(\mathcal C)$$.

Clearly, for the object $$A\oplus B$$ from $$\operatorname{Ch}(\mathcal C)$$ there are inclusion morphisms $$A\to A\oplus B\leftarrow B$$. Let $$T$$ be an object with morphisms

$$A\xleftarrow{\simeq}C_1\rightarrow T,\quad B\xleftarrow{\simeq}C_2\rightarrow T.$$

Note that it suffices to consider single-step roofs because the argument, once established, can be iterated for general roofs as above.

We see that there is a morphism $$A\oplus B\xleftarrow{\simeq} C_1\oplus C_2\to T$$, making the diagram commute. However, I fail to show its uniqueness: Given another morphism $$A\oplus B\xleftarrow{\simeq} Z\to T$$, we have to show that both are equivalent, i.e., there is an object $$Y$$ with morphisms such that

$$\begin{matrix} &&Z\\ &\swarrow&\uparrow&\searrow\\ A&\leftarrow &Y&\rightarrow &T\\ &\nwarrow&\downarrow&\nearrow\\ && C_1\oplus C_2 \end{matrix}$$

commutes, where $$Y\xrightarrow{\simeq} A$$.

Question: How to I find this object $$Y$$, showing uniqueness of the canonical morphism from $$A\oplus B$$ to $$T$$?

• the localization of an additive category is additive category.Thus you only need to check homotopy category is additive category. – Sky Dec 1 '18 at 13:57
• @Sky That $K(\mathcal C)$ is additive is clear. However, it is not totally clear to me that the localisation of an additive category yields an additive one again. – Bubaya Dec 2 '18 at 20:42

## 1 Answer

Indeed, hom sets of the derived category is easily endowed with the structure of abelian groups.

• For addition, consider two roofs $$X\xleftarrow[\simeq]{q_1}Z_1\xrightarrow{f_1} Y$$ and $$X\xleftarrow[\simeq]{q_2}Z_2\xrightarrow{f_2} Y$$. The Ore condition for a multiplicative system (which quasi-isomorphisms are an instance of) ensures that there is an object $$Z$$ and a qis $$q$$ such that $$\begin{matrix} Z&\xrightarrow[\simeq]{p_1}&Z_1\\ \llap{\scriptstyle p_2}\downarrow&&\downarrow\rlap{\scriptstyle q_1}\\ Z_2&\xrightarrow[q_2]{\simeq}&X \end{matrix}$$ commutes. Since $$q_1$$ also is a qis, the two-of-three property implies that $$p_2$$ also is a qis. Let $$q=q_1p_1=q_2p_2$$. We have brought $$f_1q_1^{-1}=(f_1p_1)q^{-1}$$ and $$f_2q_2^{-1}=(f_2p_2)q^{-1}$$ to a common denominator, such that we can simply write $$f_1q_1^{-1} + f_2q_2^{-1}=(f_1p_1 + f_2p_2)q^{-1},$$ where addition now takes place in the category $$\operatorname{Ch}(\mathcal C)$$.

• Additive inverses are immediately constructed: $$-fq^{-1}=(-1f)q^{-1}$$.

However, he original question was why the localisation has biproducts; in particular, why the universal morphism is unique. This can also be seen using the Ore condition. We just care for coproducts; the proof for products is dual.

Consider roofs $$X\xleftarrow[\simeq]{q_1}Z_1\xrightarrow{f_1} T$$ and $$X\xleftarrow[\simeq]{q_2}Z_2\xrightarrow{f_2} T$$ as above.

• There is a unique morphism $$X\oplus Y \xleftarrow[\simeq]{(q_1, q_2)} Z_1\oplus Z_2\xrightarrow{(f_1, f_2)} T$$. The two morphisms exist by the universal property of $$Z_1\oplus Z_2$$, and $$(q_1, q_2)$$ is a qis since direct sums preserve qis.
• The morphism $$X\oplus Y\leftarrow\bullet\to T$$ is unique: Let $$X\oplus Y\xleftarrow[\simeq]{p}C\xrightarrow{g} T$$ be another morphism such that $$\begin{matrix} X & \rightarrow &X\oplus Y&\leftarrow& Y\\ \uparrow && \uparrow && \uparrow\\ Z_1 && C && Z_2\\ & \searrow &\downarrow &\swarrow\\ && T \end{matrix}\label{diag1}\tag{*}$$ commutes in the derived category. By the Ore condition, we can find $$C_1$$, $$C_2$$ and morphisms such that $$\begin{matrix} C_1 & \xrightarrow{f_1} & C\\ \llap{\scriptstyle p_1}\downarrow\rlap{\scriptstyle\simeq} && \llap{\scriptstyle p}\downarrow\rlap{\scriptstyle\simeq}\\ X & \rightarrow & X\oplus Y \end{matrix}, \qquad \begin{matrix} C_2 & \xrightarrow{f_2} & C\\ \llap{\scriptstyle p_2}\downarrow\rlap{\scriptstyle\simeq} && \llap{\scriptstyle p}\downarrow\rlap{\scriptstyle\simeq}\\ Y & \rightarrow & X\oplus Y \end{matrix}\label{diag2}\tag{**}$$ commutes. Since $$C_1\oplus C_2\xrightarrow{\simeq}X\oplus Y$$, the 2-of-3 property imples that $$C\to C_1\oplus C_2$$ is a qis. Commutativity of \eqref{diag1} means that there are objects $$B_1$$, $$B_2$$ such that $$\begin{matrix} &&X&&\longrightarrow&&X\oplus Y &&\longleftarrow && Y \\ &\llap{\scriptstyle\simeq}\nearrow&\llap{\scriptstyle\simeq}\uparrow&\nwarrow\rlap{\scriptstyle\simeq}&&&\llap{\scriptstyle\simeq}\uparrow&&&\llap{\scriptstyle\simeq}\nearrow&\llap{\scriptstyle\simeq}\uparrow&\nwarrow\rlap{\scriptstyle\simeq}\\ Z_1&\xleftarrow[]{\simeq}&B_1&\xrightarrow{\simeq}&C_1 & \leftarrow & C & \rightarrow & C_2&\xleftarrow[]{\simeq}&B_2&\xrightarrow{\simeq}&Z_2\\ &\searrow&\downarrow&\swarrow&&&\downarrow&&&\searrow&\downarrow&\swarrow\\ &&T&&=&&T&&=&&T \end{matrix}$$ commutes. Finally, $$B_1\oplus B_2\xrightarrow{\simeq}Z_1\oplus Z_2$$, but also $$B_1\oplus B_2\xrightarrow{\simeq} C_1\oplus C_2\xleftarrow{\simeq} C$$, where the last qis was explained after \eqref{diag2}, and the entire diagram commutes.

The object $$B_1\oplus B_2$$ thus is the one rendering $$X\xleftarrow{\simeq} Z_1\oplus Z_2\rightarrow Y$$ and $$X\xleftarrow{\simeq} C\rightarrow Y$$ equivalent, and we have shown that the universal morphism out of $$X\oplus Y$$ is unique.