3

During a lecture my professor made an observation, she said that it can happen the following fact but she never explained how/why: let $\mathcal{A}$ a category; there are examples of morphisms $f\colon A^\bullet\to B^\bullet$ of complexes such that the morphism induced by $f$ in the derived category $D(\mathcal{A})$ is zero but the morphism induced by $f$ in the homotopy category $K(\mathcal{A})$ is not zero.

I'd like to understand how this can happen. Can anyone give me any examples please?

I'd like to precise that I know things like cone, cylinders, exact triangles.

Martin Brandenburg
  • 181,922
Grace53
  • 705
  • 5
    There can be objects which are zero in the derived category but not in the homotopy category. The identity map on such an object would be an example. – John Palmieri Dec 13 '22 at 01:41
  • 1
    @JohnPalmieri I think that what you are saying cannot be possible since the objects of the three categories $\mathcal{A}, D(\mathcal{A}), K(\mathcal{A})$ are - by definition - the same ones. Right? – Grace53 Dec 13 '22 at 02:33
  • 4
    Well, there can be objects which are isomorphic to zero in the derived category but not in the homotopy category: chain complexes which have zero homology but which are not contractible. Some examples are discussed at https://mathoverflow.net/questions/103056/when-is-an-acylic-chain-complex-contractible. – John Palmieri Dec 13 '22 at 03:34
  • 1
    $\def\C{\mathcal{C}}$The relevant result is: Let $\C$ be category, $S\subset\operatorname{Mor}\C$ be a left multiplicative system in $\C$ (in this case, there is a functor $Q:\C\to S^{-1}\C$, 04VG), and suppose $\C$ is preadditive (in this case, there is a unique preadditive structure on $S^{-1}\C$ such that $Q$ is additive, 05QD). Let $f:X\to Y$ be a morphism in $\C$. Then $Q(f)=0$ if and only if there is $s:Y\to Z$ in $S$ with $s\circ f=0$. – Elías Guisado Villalgordo Apr 12 '24 at 06:19
  • $\def\C{\mathcal{C}}$This can be deduced from 04VF (note that we can always represent a parallel pair of morphisms in $S^{-1}\C$ with common denominator thanks to 40VE). – Elías Guisado Villalgordo Apr 12 '24 at 06:25

0 Answers0