Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The prism operator for a path homotopy

Example

Let H:I×IX be a path homotopy rel {0,1} from α to β. For the identity 1-simplex ι1:Δ1I, the prism operator is the sum of the two oriented triangles cutting the square I×I along its main diagonal. If c0 and c1 are the constant singular 1-simplices at the two common endpoints, then its boundary in the unnormalized singular chain complex is PH(ι1)=βαc1+c0. The degenerate side terms vanish only after passing to normalized singular chains.

Facts & Assumptions

Given: A path homotopy H:I×IX rel {0,1} from α to β.

[L1]
[L2]

The prism operator is built by triangulating Δ1×I into two oriented 2-simplices (The prism operator of a homotopy).

[L3]

The prism operator satisfies g#f#=PH+PH (The singular chain homotopy formula).

Verification

technique · direct
1.1

By [L2], PH(ι1) is the sum of the two oriented triangles obtained from the standard triangulation of the square I×I. Applying [L3] to the singular simplex ι1 gives PH(ι1)=βαPH(ι1).

L2L3given
2.1

The chain ι1 is the terminal vertex minus the initial vertex. By [L1], these two vertices trace the constant singular 1-simplices c1 and c0, respectively, so PH(ι1)=c1c0. Substitution into step 1.1 gives PH(ι1)=βαc1+c0. Constant singular 1-simplices are genuine generators here, so they cannot be discarded in the unnormalized complex.

L1step 1.1algebra

Depends on

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