I have recently come across the following question :

Let $X$ be a (bounded below)chain complex in an arbitrary abelian category, and denote ${\sigma_{\leq n}}$ the stupid truncations functors (i.e. it replaces all $X_k$ by 0 for $k > n $). We have obvious (natural) inclusion morphisms :

$$ \sigma_{\leq n}X \hookrightarrow \sigma_{\leq n+1}X$$

and it is very easy to see that this form a diagram in the category of chain complexes such that :

$$ \mathrm{colim}\ \sigma_{\leq n}X = X.$$

My question is : does this equality still holds in the derived category?

I am not talking about homotopy colimit, but standard colimit in the derived category. More precisely, if $\gamma$ is the localization functor, I am asking if $\gamma$ preserves this colimit.

Actually, I wasn't even able to answer this question when you replace the derived category by the category of chain complexes quotiented by homotopies of chain complexes. Thus, I would already be happy with an answer to that simpler question.


1 Answer 1


No, not in general.

For example, let $R=k[x]/(x^2)$ for a field $k$ and let $X$ be the object $$\dots\stackrel{x}{\to}R\stackrel{x}{\to}R\stackrel{x}{\to}R\to0\to0\to\dots$$ of the derived category of $R$-modules, with the last non-zero term in degree zero.

To show that $X$ is not the colimit of its truncations $\sigma_{\leq n}X$ it suffices to find a non-zero map $X\to Y$ such that the restriction $\sigma_{\leq n}X\to X\to Y$ to $\sigma_{\leq n}X$ is zero for every $n$.

Take $$Y=\bigoplus_n(\sigma_{\leq n}X)[1],$$ (where $[1]$ denotes a shift to the left), and take the chain map $\alpha:X\to Y$ that in degree $n+1$ is just the map given by multiplication by $x$ from the degree $n+1$ term of $X$ to the first non-zero term of $(\sigma_{\leq n}X)[1]$.

It's easy to check that the components $X\to(\sigma_{\leq n}X)[1]$ are all homotopic to zero, but only by a chain homotopy that is non-zero in degree zero.

Each restriction to a truncation of $X$ maps into a finite subsum of $Y$, and so is homotopic to zero.

But $\alpha$ itself is not homotopic to zero (since a chain homotopy would need a map $R\to\bigoplus_nR$ in degree zero that had non-zero component for every $n$).

  • $\begingroup$ Thank you. However I do have two questions to make sure I fully understand your counter-example. First, when you say "...to the first non-zero term of $(\sigma_{\leq n}X)[1]$", didn't you mean ''...to the first non-zero term of $\bigoplus (\sigma_{\leq n}X)[1]$" instead? Secondly, at the end you have that $\alpha$ cannot be homotopic to $0$, but what we asked for is that it cannot be $0$ in the derived category. I guess you use the fact that $X$ (and $Y$) are both free modules in each degree to conclude that it's the same thing. Is that correct? $\endgroup$
    – L.Guetta
    Jun 20, 2017 at 22:32
  • 1
    $\begingroup$ @L.Guetta First question: no, I mean that, for each $n$, in degree $n+1$ the chain map $\alpha$ has component $\alpha_{n+1}$ that maps into the summand $(\sigma_{\leq n})[1]$ of $Y$. This makes sense as a chain map, since in each degree it maps into only one summand, but is not null homotopic, even though each $\alpha_n$ is, since a chain homotopy would have to involve a non-zero map in degree zero into every summand of $Y$, but a map $R\to\oplus_nR$ can only map into a finite subsum. $\endgroup$ Jun 21, 2017 at 8:29
  • $\begingroup$ @L.Guetta Second question: Yes, all complexes I consider are bounded to the right complexes of projective modules, so maps between them in the derived category and in the homotopy category are the same. $\endgroup$ Jun 21, 2017 at 8:32
  • $\begingroup$ When you say each $\alpha_n$ is null-homotopic, does it mean that you consider $\alpha_n$ as a chain map whose source is the chain complex $R$ concentrated in degree $n$? Because, it seems to me that this map is not null homotopic. $\endgroup$
    – L.Guetta
    Jun 21, 2017 at 12:23
  • $\begingroup$ @L.Guetta No, as a chain map $X\to Y$. $\endgroup$ Jun 21, 2017 at 12:37

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