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 $T$?
 A: 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.
