# Exact sequence of double complexes induces exact sequence on total complexes

This is a homework question, so I'd appreciate hints (or perhaps explanations of concepts I've not properly digested)

Anyhow: This is exercise 1.3.6 in Weibel's book on homological algebra. Let $0 \to A \to B \to C \to 0$ be an exact sequence of double complexes of modules. Show that there is a short exact sequence of total complexes, and conclude that if Tot(C) is acyclic, then $Tot(A) \to Tot(B)$ is a quasi-isomorphism.

The last part of the exercise is clear. If Tot(C) is acyclic, then the long exact sequence is of the form

$\ldots \to H_{i+1}C(=0) \to H_i(A) \to H_i(B) \to H_i(C)(=0)$

so the induced morphism on homology is an isomorphism.

The first part of the question is unclear, however. The definition of an exact sequence of double complexes is not explicitly stated, but I assume it is such that $0 \to A_{ij} \to B_{ij} \to C_{ij} \to 0$ is exact for all i,j and everything commutes.

Let $\alpha:A \to B$ be a morphism of double complexes. The induced morphism between the total complexes, $\alpha^*: Tot(A) \to Tot(B)$, is then defined, I'd assume, as $\alpha^*=d_B^h \alpha + d_B^v \alpha$ (where $d_B^h$ and $d_B^v$ denotes the horizontal and vertical differentials, respectively). The problem is how I go about showing the induced sequence is exact.

"EDIT:" After some thought, I guess a good first step would be to show that $\beta^* \circ \alpha^* = 0$, which shouldn't be too difficult. (where $\alpha^*,\beta^*$ denotes the induced morphisms of total complexes)

Edit2: I've clarified my notation a bit.

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What are alpha and beta? The morphism of total complexes is the only possible one! – Mariano Suárez-Alvarez Feb 2 '11 at 3:02
By the way, you know what a morphism of complexes is, and a double complex a complex of complexes, so you know what a morphism of double complexes is! – Mariano Suárez-Alvarez Feb 2 '11 at 3:14
As I said, I'd assumed that the definition of a morphism of total complexes was "the only" possible one. My main problem is to show that exactness is carried over to total complexes. – Fredrik Meyer Feb 2 '11 at 3:18
No, that is not what I meant. Your fifth paragraph starts with what you assume is the map between total complexes, giving a formula for alpha involving differentials and, afaict, alpha itself (!) well, that is not the map you want. – Mariano Suárez-Alvarez Feb 2 '11 at 3:54
Oh, my bad. I used the same notation for the induced map on total complexes as the original map on double complexes. I'll clarify. – Fredrik Meyer Feb 2 '11 at 4:22