# Why do we quotient by chain homotopy in the derived category.

Let $\mathcal A$ be an abelian category. To define the derived category ${\tt D}(\mathcal A)$ of $\mathcal A$ we take the category ${\tt Ch}(\mathcal A)$ of chain complexes in $\mathcal A$, quotient out null homotopic maps to get ${\tt K}(\mathcal A)$, and then invert quasi isomorphisms to get ${\tt D}(\mathcal A)$. In this definition why bother passing through ${\tt K}(\mathcal A)$?

If part of the purpose is that we are interested in the homology of these complexes more so than the specific complex representing that homology it seems weird to quotient out by only null homotopies. Why not quotient out by all maps that induce $0$ on homology (which can be strictly more than the null-homotopic ones)? Or, if that's not the purpose why quotient out by them at all? Why not just invert the quasi-isomorphisms in ${\tt Ch}(\mathcal A)$?

• One reason is that under mild hypotheses the derived category has a second description as a certain subcategory of $K(\mathcal{A})$, so it's worth defining even if all you eventually want to do is talk about the derived category. – Qiaochu Yuan Mar 10 '13 at 22:42
• @Jim: Once you invert quasi-isomorphisms, homotopic maps become equal. Thus in some sense $\mathbf{K}(\mathcal{A})$ is intermediate between $\textbf{Ch}(\mathcal{A})$ and $\mathbf{D}(\mathcal{A})$. – Zhen Lin Mar 10 '13 at 23:52
• @ZhenLin: I can see how inverting the quasi-iso's makes certain previously non-iso objects isomorphic, but it's hard to see how I can use new inverses to show that, for example, a null-homotopic map equals the zero map. Is there a quick way to see that? – Jim Mar 11 '13 at 3:41
• If you have cylinder objects $\textrm{Cyl}(M)$ (not necessarily in the sense of model categories), such that morphisms $\textrm{Cyl}(M) \to N$ correspond to homotopies of maps $M \to N$, then homotopic maps become equal in the homotopy category because the projection $\textrm{Cyl}(M) \to M$ (by definition) is a quasi-isomorphism. See this answer, for example. – Zhen Lin Mar 11 '13 at 8:12
• Excellent. Thanx for the info. – Jim Mar 11 '13 at 16:08

You could invert the quasi-isomorphisms directly. The issue is that if you invert some general class of morphisms in some arbitrary category, you have very little control over what the morphisms in the localized category will look like. For instance, the diagram

$$A = B_0 \rightarrow B'_1 \leftarrow B_1 \rightarrow B'_2 \leftarrow B_2 \rightarrow \dotsb B'_n \leftarrow B_n = B$$

in some category ${\cal C}$ defines a morphism $A \to B$ in the localized category where each map $B'_i \leftarrow B_i$ is inverted. Having to deal with these sort of diagrams all the time would be super annoying.

It turns out that the quasi-isomorphisms in the homotopy category satisfy the "Ore conditions" (and any introduction to the derived category will tell you exactly what these conditions are, so I won't bother listing them here). These conditions tell you that, once you localize and pass to the derived category, all your morphisms $A \to B$ in ${\bf D}({\cal A})$ will be of the form

$$A \rightarrow B' \leftarrow B$$

where $A \to B'$ is some map in the homotopy category ${\bf K}({\cal A})$ and $B' \leftarrow B$ is a quasi-isomorphism. Knowing this makes morphisms in ${\bf D}({\cal A})$ much easier to deal with. The quasi-isomorphisms don't satisfy these Ore conditions in the category of complexes, so without thinking about the homotopy category as an intermediate step, it would be hard to know what the morphisms in ${\bf D}({\cal A})$ look like.

By the way, my favorite introduction to the derived category is Haiman's notes here: math.berkeley.edu/~mhaiman/math256/Derived-Cat-1-5.pdf.

• Localization w.r.t. quasi-isomorphisms and the proof of Ore's conditions are well explained in the Gelfand Manin textbook on Homological Algebra. – Avitus Jun 4 '13 at 15:51

I think the point is the triangulated structure on $K(\mathcal{A})$. One can obtain $D(\mathcal{A})$ directly by localising $Ch(\mathcal{A})$ with respect to the family of all quasi-isomorphisms, since the family of all quasi-isomorphisms in $Ch(\mathcal{A})$ satisfies the axioms for a closed multiplicative system. But then it's hard to find a triangulated structure this way. Note that there is a connection between the short exact sequences in $Ch(\mathcal{A})$ and distinguished triangles (standard triangles) in ${D}(\mathcal{A})$ which helps a lot. And I don't know any other tool except distinguished triangles in a derived category that one could work with.