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 $T$?
Best 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.
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.