Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 114 1 Solution Created 2026-10-03 Updated 2026-10-14
For chain maps , a chain homotopy from to is a family of group homomorphisms satisfyingIf is a chain cycle, then is a chain boundary. Consequently , so on every homology group.
Order the vertices of the simplex as . Its simplicial chain complex hasand zero groups otherwise. An arbitrary ordering represents the sign of the permutation times the increasing ordering. Its boundary operator isIn each face obtained by deleting the vertices in positions occurs twice. Deleting first gives sign , whereas deleting first gives . These signs are opposite, so all terms cancel. The case follows directly from . Thus .
For the homology calculation, add the augmentation sending every vertex to . Regard as the group in degree of an augmented chain complex. Define the simplicial cone chain contraction byIf the simplex does not contain , expansion of its coned boundary gives . If it contains , the only face on which is nonzero is the face omitting , and . In degree zero this uses the augmentation, and in degree it says . HenceEvery positive-degree chain cycle is therefore a chain boundary. In degree zero, the same identity says , so the augmentation induces an isomorphism , with inverse . Thus, including ,This is a direct chain homotopy calculation and uses no result about cellular homology.
Simplicial cone chain contraction 2026-10-05
In an augmented chain complex of a simplex with vertices , put and when is absent, and zero when it is present. Direct expansion of the boundary operator gives . This proves that every positive-degree chain cycle is a chain boundary, without invoking cellular homology.