Usually, the derived category of some abelian category $A$ (I'm happy already with $A$-mod) is defined first taking chain complexes up to homotopy, and then localize at quasi-isomorphisms.

My question is, if one begins with $Chain(A)$=complexes in $A$ (instead of complexes up to homotopy equivalence) , and then localize at quasi-isomorphisms, do we get the same?


$$Chain(A)\overset{\tilde Q}{\to} [qis]^{-1}Chain(A)$$

the functor with the universal property w.r.t. this localization.

It is obvious that an homotopy equivalence is a quasi-isomorphism, but is it also obvious (or is it true?) that if $f\sim g$ are two homotopic maps then $Q(f)=Q(g)$? Is this category the same as the usual derived category?

  • 4
    $\begingroup$ Since homotopy equivalences are quasiisomorphisms, the localization functor $\tilde Q$ sends them to isomorphisms, hence it factors through homotopy category by its universal property. So these two localizations are the same. The advantage of localizing homotopy category is that one can use Verdier localization procedure instead of Gabriel-Zisman, which is much easier. For example, it's not obivious that the derived category will be triangulated, if you don't go through this intermediate step. $\endgroup$ Jul 10, 2020 at 1:13
  • 1
    $\begingroup$ Universal property is proved with the use of a mapping cylinder: a pair of two homotopic maps from $A$ to $B$ is the same as the single map from the mapping cylinder of an identity map of $A$ to $B$. And mapping cylinder of $id_A$ is homotopy equivalent to $A$, in two ways. $\endgroup$ Jul 10, 2020 at 1:19
  • $\begingroup$ @GrishaPapayanov You should post your first comment as an answer. $\endgroup$ Jul 10, 2020 at 2:56
  • 2
    $\begingroup$ I would add that this happens for any model category. $\endgroup$ Jul 10, 2020 at 18:02
  • $\begingroup$ Thanks Grisha and Fernando, I was looking for a direct proof ussing only mapping cones and Grisha's answer unlocked me! $\endgroup$ Jul 10, 2020 at 18:26

1 Answer 1


Turning a comment into an answer: a pair of homotopic maps $A \longrightarrow B$ is the same as a single map from a cylinder of the identity map of $A$ into $B$. This cylinder is then homotopy equivalent to $A$ in two ways, and using this one can show that any functor that sends homotopy equivalences into isomorphisms (such as localization functor $\tilde Q$) factors uniquely through the homotopy category. In other words, functor from the category of complexes into the homotopy category is a localization. It follows that quasi-isomorphism localizations of the category of complexes and of the homotopy category are equivalent.

The advantege of going through a two-step process is that the homotopy category is already triangulated, and the Verdier localization procedure for triangulated categories is somewhat simpler than the general Gabriel-Zisman. For example, it's not obivious that the derived category will be triangulated, if you won't go through this intermediate step.

The sources for that are, for example, chapter 10 of Weibel or chapter 3 of Gelfand-Manin.

  • $\begingroup$ Thanks! Actually in Gelfand-Manin the definition is localizing directly (I took the notation "Q" from the,) (after that, they do all burocracy with the homotopy category too) $\endgroup$ Jul 10, 2020 at 18:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.