Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-16/4/solution

Use the short exact sequence of coefficient groups
The groups of singular chains are free, so applying the cochain construction gives a short exact sequence of cochain complexes. Its long exact sequence in cohomology has the required terms and reduction map . Define the Bockstein homomorphism to be its connecting homomorphism.
Explicitly, represent by a mod- cocycle and lift it to an integral singular cochain . Since is a cocycle, is divisible by . Then
This is exactly the lift-and-coboundary construction for the coefficient sequence above: the mod- coboundary of lies in the image of . The quotient cochain is a cocycle, since and integral cochains have no -torsion. Replacing by another lift adds and changes the quotient by modulo ; replacing the representative by a coboundary likewise leaves its cohomology class unchanged. Thus the definition is independent of choices.
A degree-zero cocycle is a function on points constant on each path component. Choose an integer representative for its value on each component. This produces an integral zero-cocycle lift, with zero coboundary. Hence the Bockstein vanishes on for every space.
For a nonzero example in every positive degree , take the Moore space , attaching the cell by a map of mapping degree . Its positive-degree integral cellular cochain complex has the differential from degree to ; when , the preceding differential from degree zero is zero. Modulo the displayed differential vanishes, and . Lift the degree- generator to the cellular cochain with value one. Its coboundary has value , so dividing by gives the generator in degree . Naturality of cellular and singular cohomology with coefficient sequences identifies this with the Bockstein on a cyclic Moore space. Therefore is nonzero, indeed an isomorphism, in the requested degree for every prime .
Finally take integral lifts of cocycles representing and . Write and . The cochain cup product obeys the graded Leibniz identity, so
Divide by , reduce modulo and pass to cohomology. This proves the Bockstein derivation rule

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