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Direct sum and product totalisations are always isomorphic

Statement

False: Direct-sum and product totalisations are always isomorphic whenever both exist.

Facts & Assumptions

[F1]

Direct sum total complex of a double complex and Product total complex of a double complex specify the diagonal objects and total differentials.

[F2]

Countable sequence groups and tail filtrations constructs $S = (Z / 2)^{(N)}$ and $P = (Z / 2)^{N}$ and proves that they have different cardinalities.

Refutation

Given: The infinite-diagonal witness of Sum and product totalisations can differ on infinite diagonals: $C_{j, - j} = Z / 2$ for $j \geq 0$ , every other component zero, and all arrows zero.

1.1

All double-complex identities hold because all maps vanish. The degree-zero direct-sum total object is $S$ and the degree-zero product total object is $P$ by [F1, F2]. Every other total degree is zero and both total differentials are zero. Thus both totalisations exist, with infinitely many nonzero summands on their sole nonzero diagonal.

F1F2

2.1

An isomorphism of these chain complexes would induce a bijection $S \to P$ , impossible because $S$ is countably infinite and $P$ is uncountable. The canonical comparison is also explicitly nonsurjective: the constant-one sequence lies in $P$ and has infinite support, so is absent from $S$ . This verifies the failed conclusion for abstract as well as canonical isomorphisms. All zero degrees and the index $j = 0$ are included; the cardinality proof and this witness require no AC.

F2step 1.1∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Weibel, An Introduction to Homological Algebra, Chapter 5 (standard reference, not scraped)
The Stacks Project, Homological Algebra (standard reference, not scraped)