# Cohomology of double complex with exact rows

Let $(C^{p,q},d_h,d_v)$ be an unbounded double complex of modules of some algebra $A$ in an abelien category. Let $d_h,d_v$ be the horizontal and vertical differential respectively. Suppose that

(1) the horizontal rows are exact,

(2) the columns are exact except at $C^{p,0}$ for all $p$.

Is there a quick way to deduce whether the induced horizontal differential is exact on $H^i(C^{p,q},d_v)$? That is , is $H^j(H^i(C^{p,q},d_v),d_h)=0$ for all $i,j$? If not, is it possible to impose conditions on the double complex so that this is true?

I am aware that a counterexample exists if we only assume (1), see https://math.stackexchange.com/questions/1368777/double-complex-with-exact-rows, but the counterexample mentioned in the above link does not satisfy assumption (2).

If I understood correctly, this can be done by showing that the two spectral sequences associated to this double complex converges to the total complex, and that both of them collapses at page 2. Though I was wondering if one can prove this without having to resort to whole-plane spectral sequences.

Thanks!

• Try a one sided zigzag bicomplex $$K^{p,q} = \begin{cases} \mathbb{Z}, & 0\leq q=-p \text{ or } 0\leq q=-p+1; \\ 0, & \text{otherwise.} \end{cases}$$ with the differentials $d_v^{p,q}$, $d_h^{p,q}$ equal to the identity when source and target are $\mathbb{Z}$. Jan 5, 2018 at 9:04