Lemma 12.25.4. Let \mathcal{A} be an abelian category. Let K^\bullet be a complex. Let A^{\bullet , \bullet } be a double complex. Let \alpha ^ p : K^ p \to A^{p, 0} be morphisms. Assume that
For every n \in \mathbf{Z} there are only finitely many nonzero A^{p, q} with p + q = n.
We have A^{p, q} = 0 if q < 0.
The morphisms \alpha ^ p give rise to a morphism of complexes \alpha : K^\bullet \to A^{\bullet , 0}.
The complex A^{p, \bullet } is exact in all degrees q \not= 0 and the morphism K^ p \to A^{p, 0} induces an isomorphism K^ p \to \mathop{\mathrm{Ker}}(d_2^{p, 0}).
Then \alpha induces a quasi-isomorphism
K^\bullet \longrightarrow \text{Tot}(A^{\bullet , \bullet })
of complexes. Moreover, there is a variant of this lemma involving the second variable q instead of p.
Proof.
The map is simply the map given by the morphisms K^ n \to A^{n, 0} \to \text{Tot}^ n(A^{\bullet , \bullet }), which are easily seen to define a morphism of complexes. Consider the spectral sequence ({}'E_ r, {}'d_ r)_{r \geq 0} associated to the double complex A^{\bullet , \bullet }. By Lemma 12.25.3 this spectral sequence converges and the induced filtration on H^ n(\text{Tot}(A^{\bullet , \bullet })) is finite for each n. By Lemma 12.25.1 and assumption (4) we have {}'E_1^{p, q} = 0 unless q = 0 and 'E_1^{p, 0} = K^ p with differential {}'d_1^{p, 0} identified with d_ K^ p. Hence {}'E_2^{p, 0} = H^ p(K^\bullet ) and zero otherwise. This clearly implies d_2^{p, q} = d_3^{p, q} = \ldots = 0 for degree reasons. Hence we conclude that H^ n(\text{Tot}(A^{\bullet , \bullet })) = H^ n(K^\bullet ). We omit the verification that this identification is given by the morphism of complexes K^\bullet \to \text{Tot}(A^{\bullet , \bullet }) introduced above.
\square
Comments (1)
Comment #9490 by Elías Guisado on
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