# Tot and colimits

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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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 $0= \prod_n 0 = \prod_n colim_m D_{n, m} \cong colim_m \prod_n D_{n,m}= colim_m \prod_{i\leq m} A = \prod_n A$