Do levelwise quasi-isomorphisms of bicomplexes induce a quasi-isomorphism between the total complexes? Let $C^{p,q}$ be a bicomplex with differentials $d_h:C^{p,q} \to C^{p+1,q}$ and $d_v:C^{p,q} \to C^{p,q+1}$ where $d_h \circ d_v = d_v \circ d_h$. Let $D^{p,q}$ be another bicomplex defined similarly.
Assume that $C^{p,q} = D^{p,q} = 0$ if $p<0$ or $q>0$. Also, assume that there is a map of bicomplexes $f:C^{p,q} \to D^{p,q}$ which induces quasi-isomorphisms $f^{\bullet,q}:C^{\bullet,q} \to D^{\bullet,q}$ for each $q$.
1) Does $f$ induce a quasi-isomorphism $\text{Tot}^{\Pi}(f):\text{Tot}^{\Pi}(C) \to \text{Tot}^{\Pi}(D)$?
2) Does $f$ induce a quasi-isomorphism $\text{Tot}^{\oplus}(f):\text{Tot}^{\oplus}(C) \to \text{Tot}^{\oplus}(D)$?
 A: For the $\prod$ version the answer is no.
Take $C^{p,q}$ to be
$$\begin{array}{ccccccccccc}
\mathbb{Z}&\to&\mathbb{Z}&\to&0&\to&0&\to&0&\to&\dots\\
\uparrow&&\uparrow&&\uparrow&&\uparrow&&\uparrow&&\\
0&\to&\mathbb{Z}&\to&\mathbb{Z}&\to&0&\to&0&\to&\dots\\
\uparrow&&\uparrow&&\uparrow&&\uparrow&&\uparrow&&\\
0&\to&0&\to&\mathbb{Z}&\to&\mathbb{Z}&\to&0&\to&\dots\\
\uparrow&&\uparrow&&\uparrow&&\uparrow&&\uparrow&&\\
0&\to&0&\to&0&\to&\mathbb{Z}&\to&\mathbb{Z}&\to&\dots\\
\uparrow&&\uparrow&&\uparrow&&\uparrow&&\uparrow&\\
\vdots&&\vdots&&\vdots&&\vdots&&\vdots&&
\end{array}$$
where all the maps that could be are isomorphisms, and take $D^{p,q}=0$.
For the $\oplus$ version the answer is yes if your complexes are over some category where direct limits are exact (actually, if countable direct sums of exact sequences are exact, that's enough), as you can prove it by induction for the case where only the first $n$ rows are non-zero, and then take the direct limit of the truncations.  
If countable direct sums of exact sequences are not always exact, then you can easily construct a counterexample where the rows are exact sequences, and the vertical differentials are all zero.
