Solution
ID: past-exam-of-the-topology-and-groups-and-algebraic-topology-course-of-the-university-of-oxford/2007/b/7/solution
Past exam of the topology and Groups and Algebraic Topology course of the University of Oxford 2007 B 7 Solution by
Codex 0 2026-10-31
For ordered simplicial cochains, define the Alexander-Whitney cup product byusing the front and back faces of each characteristic simplex. Write for the coboundary, . Expanding the face-boundary formula gives the graded Leibniz identityIn this expansion, face omissions on the front and back portions supply the two displayed terms, and their extra terms at the shared vertex cancel with opposite signs. Hence the cup of cocycles is a cocycle. If changes by , its product with a cocycle changes by ; changing by changes it by . The product therefore descends to cohomology. It is associative by the three consecutive-face formula and has the constant degree-zero cochain as unit.
In the wedge, every positive-dimensional simplex belongs to exactly one factor. Restricting positive-degree cochains therefore gives a direct sum of the two cochain groups and splits the coboundaries in degrees above one. In degree one, the same splitting of coboundaries holds: any pair of degree-zero cochains on the factors can have their values at the glued vertices made equal by adding a constant to one factor, which does not change its coboundary. Thus restriction inducesRepresent classes from the two factors by cochains extended by zero on the other factor. If both degrees are positive, a simplex lies wholly in one factor, so one of the front or back evaluations is zero. Their cup product is already zero as a cochain. This proves all the stated mixed products vanish.
For the indicated wedge there is one unit in degree zero, two generators in degree one, and one generator in degree two; higher groups vanish. Every product of positive-degree elements is zero. Its ring is thereforeThe torus has the same additive groups, but its degree-one generators have nonzero cup product generating degree two. The two graded rings are not isomorphic. Since a homotopy equivalence induces a cohomology-ring isomorphism, the torus and this wedge are not homotopy equivalent.
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