If a morphism of pushouts of complexes (with one arrow monic) is composed of quasi-isos, then the induced arrow is one also EDIT: The original title was: If a morphism of diagrams of complexes is composed of quasi-isomorphisms, is the induced arrow a quasi-isomorphism?
Let $J$ be a small category and $C$ be the category of complexes over an abelian category. If $F,G:J\to C$ are functors and $\tau:F\Rightarrow G$ is a natural transformation such that $\tau_i:Fi\to Gi$ is a quasi-isomorphism for every $i\in J$, is $\varinjlim \tau: \varinjlim F \to \varinjlim G$ a quasi-isomorphism?
What about the case of pushouts?
This question arises to me as I try to check that the category of complexes with the subcategory of quasi-isomorphisms is a "category of weak equivalences" in the sense of Waldhausen (well, not exactly, as I'm dealing with complexes of some special exact categories, but for the present case it doesn't matter).
EDIT: The general case has been answered by Zhen Lin below, but apparently the case when one of the arrows of a pushout is monic, the result should be true. After some hours now of trying, I'm not able to do it. I've tried several things using long exact sequences of homology and also Mayer-Vietoris for complexes, but I just can't conclude. Any help would be appreciated!
 A: In general the answer is no, even for pushouts. The whole point of homotopy colimits is to fix this deficiency of ordinary colimits.
For example, let $D$ be the two-term chain complex with $D_0 = \mathbb{Z} \oplus \mathbb{Z}$, $D_1 = \mathbb{Z}$ and differential given by $1 \mapsto (1, -1)$. Consider the following pushout diagram,
$$\begin{array}{ccc}
\mathbb{Z} \oplus \mathbb{Z} & \rightarrow & \mathbb{Z} \\
\downarrow & & \downarrow \\
D & \rightarrow & S
\end{array}$$
where $\mathbb{Z} \oplus \mathbb{Z} \to D$ is given by $\mathrm{id}$ in degree 0 and $\mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z}$ is given by $(x, y) \mapsto x + y$. Then $S$ is a two-term chain complex with $S_0 = S_1 = \mathbb{Z}$ and zero differential. On the other hand, the morphism $D \to \mathbb{Z}$ given by $(x, y) \mapsto x + y$ in degree 0 is a quasi-isomorphism, and the corresponding pushout diagram is
$$\begin{array}{ccc}
\mathbb{Z} \oplus \mathbb{Z} & \rightarrow & \mathbb{Z} \\
\downarrow & & \downarrow \\
\mathbb{Z} & \rightarrow & \mathbb{Z}
\end{array}$$
but $S$ is not quasi-isomorphic to $\mathbb{Z}$.
A: The diagram chase I had tried missed only one ingredient: the fact the cokernels of a pair of parallel arrows in a pushout are compatibly isomorphic. I didn't know this fact, actually. It's straightforward to prove using the characterization of a pushout by an appropriate right exact sequence.
Here's the proof, which I wrote up in $\LaTeX$:



