LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-08-31

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The three-cone calculation for a composite chain map

Statement

For composable chain maps $C_{∙} \overset{f}{\to} D_{∙} \overset{g}{\to} E_{∙},$ let $α : Cone (f) \to Cone (g f), β : Cone (g f) \to Cone (g)$ be the cone maps induced by the strict squares $(1_{C}, g)$ and $(f, 1_{E})$ . Then $Cone (α)$ is chain-isomorphic to $Cone (g) \oplus Cone (1_{C [1]}),$ hence chain-homotopy equivalent to $Cone (g)$ .

Facts & Assumptions

Given: Composable chain maps $C_{∙} \overset{f}{\to} D_{∙} \overset{g}{\to} E_{∙}$ .

[L1]

A strict square of chain maps induces a chain map of cones (A morphism of chain maps induces a chain map of cones).

[L2]

The cone differential is $d (y, x) = (d (y) + f (x), - d (x))$ with the relevant map in the upper-right corner (The mapping cone of a chain map).

[L3]

The cone of an identity map is contractible (The cone of an identity map is contractible).

Proof

technique · direct

1.1L1L2givenconstructalgebra

By [L1], the map $α$ is $α_{n} (d, c) = (g_{n} (d), c) .$ Writing out the cone differential from [L2] shows that $Cone (α)_{n} = E_{n} \oplus C_{n - 1} \oplus D_{n - 1} \oplus C_{n - 2}$ . The change of coordinates $Θ_{n} (e, c, d, c^{'}) : = ((e, d + f_{n - 1} (c)), (c, c^{'}))$ is a degreewise isomorphism from $Cone (α)_{n}$ to $Cone (g)_{n} \oplus Cone (1_{C [1]})_{n}$ .

2.1L3step 1.1algebra∎

A direct substitution into the differential formulas shows that $Θ$ is a chain map. Therefore $Cone (α)$ is chain-isomorphic to $Cone (g) \oplus Cone (1_{C [1]})$ . The second summand is contractible by [L3], so the direct sum is chain-homotopy equivalent to $Cone (g)$ .

The three-cone calculation for a composite chain map

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources