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.