When one wants to prove that a morphism $f_*$ between two chain complexes $\left(C_*\right)$ and $\left(D_*\right)$ is zero in homology, one of the standard approaches is to look for a chain homotopy, i. e., for a map $U_n:C_n\to D_{n+1}$ defined for every $n$ that satisfies $f_n=d_{n+1}U_n+U_{n-1}d_n$ for every $n$. However, this is not strictly necessary: For example, it is often enough to have two maps $U_n:C_n\to D_{n+1}$ and $V_n:C_n\to D_{n+1}$ defined for every $n$ that satisfy $f_n=d_{n+1}U_n+V_{n-1}d_n$ for every $n$. This way, when constructing $U_n$ and $V_n$, one doesn't have to care for them to "fit together", because each is used only one time.

However, at least my experience suggests that one does not gain much from this - when one tries to construct these $U_n$ and $V_n$, they turn out (after some simplification) to be the same.

My question is: What is the deeper reason behind this? Why do chain homotopies like to "fit together" although they don't need to?

I am sorry if this makes no sense...

EDIT: Thanks David, it seems I can't write a single absatz without a stupid mistake.

  • $\begingroup$ Writing f=du+ud shows that f is zero on homology, not that it is an isomorphism. Interesting question, I'll see if I can think of more to say. $\endgroup$ Commented Dec 15, 2009 at 12:42
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    $\begingroup$ I don't think this is a "soft question", so I'm gonna remove the soft-question tag. $\endgroup$ Commented Dec 15, 2009 at 19:26
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    $\begingroup$ I think this is related to the fact in topology that a map $f\colon X\to Y$ is a homotopy equivalence iff there are maps $g,h\colon Y\to X$ so that $f\circ g \simeq id_Y$ and $h\circ f \simeq id_X$. In practice you like for $g$ and $h$ to be the same map, but they don't need to be. I haven't worked out the rest of the details as to how exactly this relates to your question about chain complexes, so I might be drawing this comparison in error. I'd be interested to know one way or the other. $\endgroup$ Commented Apr 12, 2010 at 15:00

4 Answers 4


EDITED, because I think I see the big picture now.

We have the following theorem: let $C$ and $D$ be complexes of projective objects. The following are equivalent: (a) the map $f: C \to D$ is the zero map in the derived category (b) there is a homotopy between $f$ and $0$.

In this theorem, the definition of homotopy is that $f=du+ud$. So my answer to your question is: In practice, if a map induces the zero map on homology, it is probably zero in the derived category. There do exist maps which are of the form $du+vd$, but are not of the form $du+ud$, see my other answer.

Paragraph about hereditary algebras deleted because I don't think it was quite right.

Some more elementary observations

(1) It is required that $f$ be a map of chain complexes, so $df=fd$. So we want $d(du+vd) =(du+vd)d$ or $dud=dvd$. This doesn't force $u=v$, but it is the easiest way to achieve it.

(2) There is a topological way of thinking of the condition $du+ud=f$, which I learned from Joel Kamnitzer. Let $I$ be the chain complex with $I_1=\mathbb{Z}$, $I_0 = \mathbb{Z}^2$ and the map $I_1 \to I_0$ given by $(1 \ -1)$. Let $\partial I$ be the subchain complex where $(\partial I)\_0=I_0$ and $(\partial I)\_i=0$ for all other $i$.

Then writing $f=du+ud$ is equivalent to finding a map $u:C \times I \to D$ such that, when we restrict to $C \times (\partial I)$, we have the map $f$ on one component and $0$ on the other. $I$ is the chain complex of the obvious triangulation of the unit interval. Thinking of $I$ as the unit interval, this really is a homotopy between $f$ and $0$. I can't think of an analogous geometric motivation for $f=du+vd$.

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    $\begingroup$ Minor comment: You need to assume something like the domain complex being bounded below (or having an exhaustive filtration with subquotients being bounded-below complexes of projectives) in order to be able to show that a map f which is zero in the derived category implies the existence of a homotopy from f to 0. $\endgroup$ Commented Dec 15, 2009 at 17:19
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    $\begingroup$ @Tyler: complexes are bounded on one side. @David: The top half of this solution together with the counterexample in the other answer are just a really nice answer. I learned stuff from it. $\endgroup$ Commented Dec 15, 2009 at 18:23
  • $\begingroup$ Thanks Greg! For completeness, I'm going to add one more counter-example to the other answer. $\endgroup$ Commented Dec 15, 2009 at 18:53
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    $\begingroup$ @Greg: It's been my understanding that this isn't convention, it's usually specified if necessary? There is a perfectly good derived category of unbounded complexes. $\endgroup$ Commented Dec 15, 2009 at 19:07
  • $\begingroup$ Sorry, I meant WELL-BEHAVED complexes are generally bounded on one side. I dropped part of what I wanted to say. $\endgroup$ Commented Dec 15, 2009 at 19:30

Here is an example of a map of chain complexes which can be realized as $du+vd$ but not as $du+ud$.

Let $C$ and $D$ both be the chain complex

$$\cdots 0 \to \mathbb{Z}/p^2 \to \mathbb{Z}/p^2 \to 0 \to \cdots$$

where the nontrivial map is multiplication by $p$.

Map $C$ to $D$ by multiplication by $p$ in the first nontrivial degree, and by zero in the other nontrivial degree. This is $du+vd$, where $u$ is $1$ and $v$ is $0$ (in the only nontrivial degree).

On the other hand, this map cannot be written as $du+ud$. If we had such a representation, then $u$ would be multiplication by $a$ for some $a \in \mathbb{Z}/p^2$. The two vertical maps would then both be $pa$. In particular, it is impossible that one would be zero and the other not.

If we consider these complexes as objects in the derived category of $(\mathbb{Z}/p^2)$-modules, they are complexes of projective objects. So this is an example of a map of complexes which induces zero maps on cohomology but is not zero in the derived category.

What happens if we pass to a quotient category where maps like this are zero? I have no idea! Does anyone?

For completeness, I add an example of a map which induces zero on homology and can not be written as $du+vd$. Consider $$\begin{matrix} 0 & \to & \mathbb{Z} & \to & \mathbb{Z} \\\\ \downarrow & & \downarrow & & \downarrow \\\\ \mathbb{Z} & \to & \mathbb{Z} & \to & 0 \end{matrix}$$

where the nontrivial horizontal maps are multiplication by $p$, and the vertical map is the identity. The nontrivial homology is in different degrees in top and bottom, so the map on homology is zero. But any map of the form $du+vd$ would be $0$ modulo $p$ in the central column.

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    $\begingroup$ I think the point is that a true chain homotopy preserves not on homology, but also Bockstein homomorphisms. $\endgroup$ Commented Dec 15, 2009 at 15:19
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    $\begingroup$ In the last sentence, do you mean "0 mod p"? $\endgroup$ Commented Dec 15, 2009 at 19:19
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    $\begingroup$ Greg, what do you mean by "preserving Bockstein homomorphisms"? At first, are they defined at all? The definition I know involves tensoring an exact sequence of abelian groups with $C_n$ for every $n$, hoping that this will produce another short exact sequence; however, this requires the $C_n$ to be flat. $\endgroup$ Commented Dec 16, 2009 at 18:47
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    $\begingroup$ Still, even when the Bockstein homomorphisms are well-defined, what does "preserving" them mean? Aren't they canonical and therefore always preserved by a map of chain complexes? $\endgroup$ Commented Dec 16, 2009 at 19:50
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    $\begingroup$ Darij: You're right that my remark and my thinking were both muddled. What I should have said is that, in David's examples, your generalized chain homotopies are no longer chain homotopies if you tensor the complexes with $\mathbb{Z}/p$; they create non-zero homology maps in the new coefficients. My intuition is that their behavior after tensoring can be expressed in terms of the Bockstein maps that become available. There is a "Bockstein spectral sequence", related to the universal coefficient theorem, that may help clarify the situation. $\endgroup$ Commented Dec 18, 2009 at 6:58

Amusingly, I have considered the category of complexes where maps of the form du+vd are killed in: Weight structures vs. $t$-structures; weight filtrations, spectral sequences, and complexes (for motives and in general), to appear in J. of K-theory, http://arxiv.org/abs/0704.4003, in section 3.1. My main observation is that the composition of two chain complex morphisms of form du+vd yield a morphism homotopic to zero; so you actually do not forget very much information. In particular, the projection functor from the homotopy category of complexes to this more coarse category is conservative i.e. a non-isomorphism does not become an isomorphism.

You may also find some motivation for introducing such a category in Remark 1.5.2 of my paper.

  • $\begingroup$ That's a very nice result! So if $du+vd$ and $dU+Vd$ are two chain morphisms, then $\left(du+vd\right)\left(dU+Vd\right)=ds+sd$, where $s=vdU+uVd$ (because $dud=dvd$ and $dUd=dVd$ due to $du+vd$ and $dU+Vd$ being chain morphisms). I wish I would understand some of your paper, but I fear I'll have to learn some K-theory first. $\endgroup$ Commented Jan 5, 2010 at 23:54
  • $\begingroup$ My paper is certainly related to K-theory; yet none of K-theory is necessary to understand it. It could be interesting to you if you understand t-structures. $\endgroup$ Commented Jan 6, 2010 at 16:54

I have just found another article on this: Absolute Homology by Michael Barr.


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