TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06

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Subdivision is chain homotopic to the identity

Statement

The operators $S$ and $1$ obey $1 - S = \partial T + T \partial$ .

Facts & Assumptions

Given: The recursive prism operator $T$ and chain map $S$ .

Proof

technique · induction

1.1

In dimension $0$ , $S = 1$ and $T = 0$ . Assume the identity on all faces of an $n$ -simplex $σ$ .

givenbaseassume-hyp

2.1

On the universal simplex, the induction hypothesis applied to $\partial ι_{n}$ gives $\partial (ι_{n} - P (\partial ι_{n})) = S (\partial ι_{n}) .$ The affine-cone formula and the recursive definition of subdivision therefore give $\partial P_{n} = ι_{n} - P (\partial ι_{n}) - b_{n} S (\partial ι_{n}) = ι_{n} - P (\partial ι_{n}) - S (ι_{n}) .$

step 1.1ihalgebra

3.1

Pushing the equality in step 2.1 forward along $σ : Δ^{n} \to X$ gives $\partial T σ + T \partial σ = σ - S σ$ . Linearity proves the identity on chains.

step 2.1discharge-induction∎

Depends on

Used by

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Allen Hatcher, Algebraic Topology, Proposition 2.21 (standard reference, not scraped)