For $\mathcal{S}$ the $(\infty,1)$-category of spaces [its homotopy category $h\mathcal{S}$ does not have pushouts or pullbacks](https://mathoverflow.net/questions/239383/the-homotopy-category-is-not-complete-nor-cocomplete). Even if it does, they won't always agree with the (homotopy) pushouts or pullbacks in $\mathcal{S}$.


Generally, filtered colimits are much better behaved than general ones. For example [filtered colimits in the (Quillen) model category $\mathit{sSet}$ compute homotopy colimits in $\mathcal{S}$.](https://mathoverflow.net/questions/235526/when-do-colimits-agree-with-homotopy-colimits) 
In light of this I was wondering the following:



> 
> Does $h\mathcal{S}$ have filtered colimits and, if so, does the functor $\mathcal{S} \to h\mathcal{S}$ preserve them?
> 
> 
> 


More generally one could ask the same for any presentable $(\infty,1)$-category $\mathcal{C}$; I particularly care about the case of the derived category $\mathcal{D}(R)$ of a ring $R$.



Can filtered colimits be computed in the homotopy category?
No, for a diagram $X: I \to \mathcal{S} \to h\mathcal{S}$ the colimit in $h\mathcal{S}$ would satisfy $[\mathrm{colim} X(i),Y] \cong \lim [X(i),Y]$ where brackets denote morphisms in $h\mathcal{S}$. Informally, the set of maps in $h\mathcal{S}$ from the (homotopy) colimit in $\mathcal{S}$ should involve also higher derived limits.


For a concrete example, you could take $I = (\mathbb{N},<)$, $X(i) \simeq S^1$ for all $i$, $X(i) \to X(i+1)$ a map of degree 2. Then $\mathrm{colim}(X(i)) \simeq K(\mathbb{Z}[1/2],1)$ is not the colimit in $h\mathcal{S}$ since it has distinct morphisms to $Y = K(\mathbb{Z},2)$ which become equal upon restricting to each $X(i)$.


Similar examples exist in $D(\mathbb{Z})$.


