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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 $T$?

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  • $\begingroup$ the localization of an additive category is additive category.Thus you only need to check homotopy category is additive category. $\endgroup$ – Sky Dec 1 '18 at 13:57
  • $\begingroup$ @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. $\endgroup$ – Bubaya Dec 2 '18 at 20:42
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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.

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