Let $C$ be the category of chain complexes over a field $F$ and $C^\prime$ be the subcategory of chain complexes with zero differentials. If $X:I\to C$ is a functor, there is an induced "homology" functor $HX:I\to C^\prime$.
Now $X$ and $HX$ are levelwise homotopy equivalent (because we are working over a field, see for example here for details) but not naturally, i.e., there is no morphism $X\to HX$ (or the other way) in the functor category $C^I$. But I can view $X$ and $HX$ in the category $\operatorname{Coh}(I,C)$ of homotopy coherent diagrams and in this category (I think) there is a morphism between them (i.e., a h.c. transformation).
By work of Cordier and Porter (building on Vogt's work), the transformation is invertible (up to homotopy) because it is a levelwise equivalence. The homotopy colimit is a functor from $\operatorname{Ho}(\operatorname{Coh}(I,C))$ to $\operatorname{Ho}(C)$ so $X$ and $HX$ have equivalent homotopy colimits.
Question: If $I$ has simplicial shape, we are talking about double complexes and totalization is a model for the homotopy colimit. But the double complex spectral sequence collapses in the case of $HX$, while the one for $X$ need not, so how can the homotpy colimits be equivalent?
Best Answer
The answer is that the double complex does not collapse in the case of $HX$ either. The issue is that you have a coherent simplicial diagram valued in chain complexes with differential zero, rather than a strict one. Such a coherent diagram includes data such as chain homotopies expressing equivalence of double composites, etc, etc, and this data has an impact on the homotopy colimit.
For example, a coherent simplicial object has a chain homotopy $h$ from $d^2$ to 0 as maps from from $C_2$ to $C_0$. If $x$ is a cycle in the kernel of $d$, then $h(x)$ is a cycle in $C_0$ of degree one higher; that's an expression of $d_2(x)$.