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.



A simple colimit in the derived category?
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 $n$).


