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Let $(K^{p,q},\delta,d)$ be a double complex of modules. We assume that $\delta$ of degree $(1,0)$, $d$ has degree $(0,1)$ and $d$ and $\delta$ commute.

Since $d$ and $\delta$ commute, then $\delta$ induces a differential operator on $H_d(K)$ by $\delta[\omega]=[\delta\omega]$ (where $[\omega]$ denotes the cohomology class of $\omega\in K$ for $d$).

I am considering the following statament:

If the rows of the double complex are exact (that is $\operatorname{Im}(\delta)= \operatorname{Ker}(\delta)$), then $H_d(K)$ is also exact.

Trying to prove this statement gave me the intuition that it is probably false. But my lack of experience in homological algebra makes that I cannot construct a counter-example. Could someone provide a counter-example to the above statement?

In addition, are there simple conditions that makes the previous statement true?

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2 Answers 2

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This is not necessarily true. Consider the following bicomplex: $$K^{p,q} = \begin{cases} \mathbb{Z}, & q=-p \text{ or } q=-p+1; \\ 0, & \text{otherwise.} \end{cases}$$ enter image description here

The differentials are defined as follows: $d : K^{p,-p} \to K^{p,-p+1}$ (the vertical maps) is multiplication by $2$, while $\delta : K^{p,-p} \to K^{p+1,-p}$ (the horizontal maps) is the identity (all the other ones are necessarily zero). It's clear that the rows are exact, since $\delta$ is the identity on the only possible term. But the cohomology $H(K,d)$ is concentrated just above the diagonal with: $$H^{p,q}(K,d) = \begin{cases} \mathbb{Z}/2\mathbb{Z}, & q=-p+1; \\ 0, & \text{otherwise.} \end{cases}$$ And the rows are not exact (the map induced by $\delta$ is zero).


A prominent feature of this bicomplex is that it is not bounded: in total degrees $p+q=0$ and $p+q=1$, there is an infinite number of nonzero terms. If the bicomplex were bounded, I think the situation could be different.

(If you don't know anything about spectral sequences, you can ignore this paragraph)
Your condition on $\delta$ says that the first page of the spectral sequence associated to the bicomplex $(K,d,\delta)$ vanishes. By standard arguments, the spectral sequence associated to the bicomplex $(K,\delta,d)$ (with switched order for the differentials) abuts to the same $E^\infty$ page as the first spectral sequence when the complex is bounded, though I don't know if one necessarily has that $E^1$ of the second sequence vanishes when $E^1$ of the first sequence vanishes – in fact, I rather doubt it.

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  • $\begingroup$ If you restrict to $r$ of the rows, the $(K,\delta,d)$ spectral sequence has a $d_r$ and its $E^{r+1}$ page vanishes. $\endgroup$ Commented Jul 21, 2015 at 16:00
  • $\begingroup$ Ah yes @EricWofsey of course. I'm too used to these things degenerating after a page or two... $\endgroup$ Commented Jul 21, 2015 at 16:01
  • $\begingroup$ @NajibIdrissi I am sorry for bringing up an old post, but what about for a double complex with exact rows and 1 non-exact column $(H(K^p,q,\delta)=0$ for all $p$ and $H(K^p,q,d)=0$ for all $q$ except $q=0$)? This is stronger than OP's assumption where your counter example does not apply. I was wondering if it's still true that the induced $\delta$ on $H(K,d)$ is exact. Thanks! $\endgroup$
    – Steve
    Commented Jan 2, 2018 at 4:06
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I would try to prove that the result is still false also in the case of a bounded double complex.

Let's work in an abelian category $A$ with enough projectives, and let $F$ be a right-exact functor (for instance, you could think to the tensor functor over the category of modules). Let $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0 $ be a ses such that $0 \rightarrow F(A) \rightarrow F(B) \rightarrow F(C) \rightarrow 0$ is not exact.

Choose now two projective resolutions: $P.$, $R.$ respectively for $A$ and $C$. Thanks to the horseshoe lemma there exists a projective resolution of $B$, call it $Q.$, such that extends to a short sequence (in $Ch(A)$) $0 \rightarrow P. \rightarrow Q. \rightarrow R. \rightarrow 0$

Because $R.$ is made by projective objects, the previous ses splits leading to the short exact sequence $0 \rightarrow F(P.) \rightarrow F(Q.) \rightarrow F(R.) \rightarrow 0$ - call it $(*)$ (recall that any additive functor - a condition assumed on the definition of right-exactness - sends split ses to split ses).

Now the ground is ready for making the double complex simply putting on the zero-row the zero-level of $(*)$, on the first raw the level 1 of $(*)$, and so on (you probably need to reverse the arrows).

Observe that the double complex obtained is bounded, exact in the rows, but not in the columns. After having taken the homology respect to the vertical direction, you will obtain the zero object everywhere except in the zero row, where there will be the sequence $0 \rightarrow F(A) \rightarrow F(B) \rightarrow F(C) \rightarrow 0$ (recalling the definition of left derived functor, and the fact that it coincides with the original functor on the zero degree) that is not exact by hypothesis.

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