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This must be a well-known exercise with spectral sequences, but I don't know a reference for it. I'm trying to figure out when does $Tot$ commute with colimits.

More precisely, let $X$ be a double cochain complex of, say, $R$-modules, $R$ a commutative ring with unit, or, more generally, a double complex in an abelian category. Let $\cal{C}$ denote the category of these double cochain complexes.

We have two different total functors, $\mbox{Tot}^\prod$ and $\mbox{Tot}^{\bigoplus}$, from the category of double complexes to the category of cochain complexes:

$$ \mbox{Tot}^{\prod}(X)^n = \prod_{p+q=n}X^{p,q} \qquad \mbox{and} \qquad \mbox{Tot}^{\bigoplus}(X)^n = \bigoplus_{p+q=n}X^{p,q} \quad . $$

Let $\mbox{Tot}$ denote anyone of them and let $I$ be a (filtered) category, and $X: I \longrightarrow \cal{C}$ a functor. We have a natural morphism

$$ \theta: \varinjlim_i \mbox{Tot} (X_i) \longrightarrow \mbox{Tot} (\varinjlim_i X_i) \quad . $$

When dealing with $\mbox{Tot}^\bigoplus$, this $\theta$ is an isomorphism, because a direct sum is a colimit and colimits commute with colimits.

What happens when we take $\mbox{Tot}^\prod$? Is $\theta$ at least a quasi-isomorphism (a morphism inducing an isomorphism in cohomology)? In which cases? Do we need some extra hypothesis on the abelian category (AB...)? Is the hypothesis "filtered" really needed, or we can deal with arbitrary colimits in general?

Of course, if our double complex has finite diagonals, then $\mbox{Tot}^\prod = \mbox{Tot}^\bigoplus$, and we are done. But what happens without this hypothesis?

I'm mainly interested in the case of a right half-plane double complex, that is $X^{p,q} = 0$ if $p<0$, but I'll be glad to learn about all possible cases.

Any references or hints will be welcome.

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up vote 8 down vote accepted

Imagine that all double complexes in the image of your functor X: I → C have both differentials equal to zero. Moreover, all terms of these bicomplexes outside of a fixed diagonal are also zero. Then you are asking, quite simply, whether colimits commute with countable products. If they don't, your morphism θ cannot be a quasi-isomorphism (being a non-isomorphism of complexes with zero differentials). And of course, if in a certain abelian category countable filtered colimits commute with countable products (and both exist), then all objects of this category are zero.

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Sorry, but why are necessarily all the objects zero? – a.r. Jun 16 '10 at 0:56
Consider the set of diagrams D_n = (0->0->...->A->A->...) -- 0 on the first n positions and A on the subsequent ones, where A is a certain fixed object in our abelian category, and the maps between copies of A are the identity maps. If the countable filtered colimit commutes with the countable product for this set of diagrams, it simply means that the natural map from the coproduct of a countable set of copies of A to their product is an isomorphism. – Leonid Positselski Jun 16 '10 at 8:32
It is a standard fact (mentioned in Grothendieck's Tohoku paper) that in this case the object A must be zero. Basically, the isomorphism of the finite coproduct and the finite product in an additive category allows to add morphisms, and an isomorphism between the countable coproduct and product would allow to take countable sums of morphisms. In particular, there is a well-defined countable sum of copies of the identity endomorphism of A. This can be only non-contradictory when the identity endomorphism is zero. – Leonid Positselski Jun 16 '10 at 8:36
Ok, thank you very much! – a.r. Jun 16 '10 at 8:40

I seem that instead have to consider the the diagram $D_n=(A\to A....\to A\to 0\to 0\to ...)$ definited as the yours but reciprocally changing $A$ by $0$, then If the countable filtered colimit commutes with the countable product for this set of diagrams you have that $0= \prod_n 0 = \prod_n colim_m D_{n, m} \cong colim_m \prod_n D_{n,m}= colim_m \prod_{i\leq m} A = \prod_n A$

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