Remark 12.18.7. Let $\mathcal{A}$ be an additive category with countable direct sums. Let $\text{DoubleComp}(\mathcal{A})$ denote the category of double complexes. We can consider an object $A^{\bullet , \bullet }$ of $\text{DoubleComp}(\mathcal{A})$ as a complex of complexes as follows

$\ldots \to A^{-1, \bullet } \to A^{0, \bullet } \to A^{1, \bullet } \to \ldots$

For the variant where we switch the role of the indices, see Remark 12.18.6. In this remark we show that taking the associated total complex is compatible with all the structures on complexes we have studied in the chapter so far.

First, observe that the shift functor on double complexes viewed as complexes of complexes in the manner given above is the functor $[1, 0]$ defined in Remark 12.18.5. By Remark 12.18.5 the functor

$\text{Tot} : \text{DoubleComp}(\mathcal{A}) \to \text{Comp}(\mathcal{A})$

is compatible with shift functors, in the sense that we have a functorial isomorphism $\gamma : \text{Tot}(A^{\bullet , \bullet }) \to \text{Tot}(A^{\bullet , \bullet }[1, 0])$.

Second, if

$f, g : A^{\bullet , \bullet } \to B^{\bullet , \bullet }$

are homotopic when $f$ and $g$ are viewed as morphisms of complexes of complexes in the manner given above, then

$\text{Tot}(f), \text{Tot}(g) : \text{Tot}(A^{\bullet , \bullet }) \to \text{Tot}(B^{\bullet , \bullet })$

are homotopic maps of complexes. Indeed, let $h = (h^ p)$ be a homotopy between $f$ and $g$. If we denote $h^{p, q} : A^{p, q} \to B^{p - 1, q}$ the component in degree $p$ of $h^ q$, then this means that

$f^{p, q} - g^{p, q} = d_1^{p - 1, q} \circ h^{p, q} + h^{p + 1, q} \circ d_1^{p, q}$

The fact that $h^ p : A^{p, \bullet } \to B^{p - 1, \bullet }$ is a map of complexes means that

$d_2^{p - 1, q} \circ h^{p, q} = h^{p, q + 1} \circ d_2^{p, q}$

Let us define $h' = ((h')^ n)$ the homotopy given by the maps $(h')^ n : \text{Tot}^ n(A^{\bullet , \bullet }) \to \text{Tot}^{n - 1}(B^{\bullet , \bullet })$ using $h^{p, q}$ on the summand $A^{p, q}$ for $p + q = n$. Then we see that

$d_{\text{Tot}(B^{\bullet , \bullet })} \circ h' + h' \circ d_{\text{Tot}(A^{\bullet , \bullet })}$

restricted to the summand $A^{p, q}$ is equal to

$d_1^{p - 1, q} \circ h^{p, q} + (-1)^{p - 1} d_2^{p - 1, q} \circ h^{p, q} + h^{p + 1, q} \circ d_1^{p, q} + h^{p, q + 1} \circ (-1)^ p d_2^{p, q}$

which evaluates to $f^{p, q} - g^{p, q}$ by the equations given above. This proves the second compatibility.

Third, suppose that in the paragraph above we have $f = g$. Then the assignment $h \leadsto h'$ above is compatible with the identification of Lemma 12.14.9. More precisely, if we view $h$ as a morphism of complexes of complexes $A^{\bullet , \bullet } \to B^{\bullet , \bullet }[-1, 0]$ via this lemma then

$\text{Tot}(A^{\bullet , \bullet }) \xrightarrow {\text{Tot}(h)} \text{Tot}(B^{\bullet , \bullet }[-1, 0]) \xrightarrow {\gamma ^{-1}} \text{Tot}(B^{\bullet , \bullet })[-1]$

is equal to $h'$ viewed as a morphism of complexes via the lemma. Here $\gamma$ is the identification of Remark 12.18.5. The verification of this third point is immediate.

Fourth, let

$0 \to A^{\bullet , \bullet } \to B^{\bullet , \bullet } \to C^{\bullet , \bullet } \to 0$

be a complex of double complexes and suppose we are given splittings $s^ p : C^{p, \bullet } \to B^{p, \bullet }$ and $\pi ^ p : B^{p, \bullet } \to A^{p, \bullet }$ of this as in Lemma 12.14.10 when we view double complexes as complexes of complexes in the manner given above. This on the one hand produces a map

$\delta : C^{\bullet , \bullet } \longrightarrow A^{\bullet , \bullet }[0, 1]$

by the procedure in Lemma 12.14.10. On the other hand taking $\text{Tot}$ we obtain a complex

$0 \to \text{Tot}(A^{\bullet , \bullet }) \to \text{Tot}(B^{\bullet , \bullet }) \to \text{Tot}(C^{\bullet , \bullet }) \to 0$

which is termwise split (see below) and hence comes with a morphism

$\delta ' : \text{Tot}(C^{\bullet , \bullet }) \longrightarrow \text{Tot}(A^{\bullet , \bullet })$

well defined up to homotopy by Lemmas 12.14.10 and 12.14.12. Claim: these maps agree in the sense that

$\text{Tot}(C^{\bullet , \bullet }) \xrightarrow {\text{Tot}(\delta )} \text{Tot}(A^{\bullet , \bullet }[1, 0]) \xrightarrow {\gamma ^{-1}} \text{Tot}(A^{\bullet , \bullet })$

is equal to $\delta '$ where $\gamma$ is as in Remark 12.18.5. To see this denote $s^{p, q} : C^{p, q} \to B^{\bullet , q}$ and $\pi ^{p, q} : B^{p, q} \to A^{p, q}$ the components of $s^ q$ and $\pi ^ q$. As splittings $(s')^ n : \text{Tot}^ n(C^{\bullet , \bullet }) \to \text{Tot}^ n(B^{\bullet , \bullet })$ and $(\pi ')^ n : \text{Tot}^ n(B^{\bullet , \bullet }) \to \text{Tot}^ n(A^{\bullet , \bullet })$ we use the maps whose components are $s^{p, q}$ and $\pi ^{p, q}$ for $p + q = n$. We recall that

$(\delta ')^ n = (\pi ')^{n + 1} \circ d_{\text{Tot}(B^{\bullet , \bullet })}^ n \circ (s')^ n : \text{Tot}^ n(C^{\bullet , \bullet }) \to \text{Tot}^{n + 1}(A^{\bullet , \bullet })$

The restriction of this to the summand $C^{p, q}$ is equal to

$\pi ^{p + 1, q} \circ d_1^{p, q} \circ s^{p, q} + \pi ^{p, q + 1} \circ (-1)^ p d_2^{p, q} \circ s^{p, q} = \pi ^{p + 1, q} \circ d_1^{p, q} \circ s^{p, q}$

The equality holds because $s^ p$ is a morphism of complexes (with $d_2$ as differential) and because $\pi ^{p, q + 1} \circ s^{p, q + 1} = 0$ as $s$ and $\pi$ correspond to a direct sum decomposition of $B$ in every bidegree. On the other hand, for $\delta$ we have

$\delta ^ p = \pi ^ p \circ d_1 \circ s^ p : C^{p, \bullet } \to A^{p + 1, \bullet }$

whose restriction to the summand $C^{p, q}$ is equal to $\pi ^{p + 1, q} \circ d_1^{p, q} \circ s^{p, q}$. Thus we get the same as before which matches with the fact that the isomorphism $\gamma : \text{Tot}(A^{\bullet , \bullet }) \to \text{Tot}(A^{\bullet , \bullet }[1, 0])$ is defined without the intervention of signs.

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