Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-22/1/c/ii/solution
Past exam of the mathematics course of the University of Cambridge 2005 iii Paper 22 1 c ii Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-14
A stable map induces a degree-preserving additive natural transformation of represented cohomology theories by postcomposition with in degree . This transformation commutes with the suspension isomorphisms. The resulting mapis onto, but need not be one-to-one. Its kernel consists of hyperphantom maps of spectra: maps for which for every stable mapfor every based space and every . Equivalently, they induce zero in all degrees of represented cohomology theory on all based spaces. ThereforeThis condition tests arbitrary suspension spectra with negative shifts. Merely testing finite spectra defines a phantom map of spectra, which is a weaker condition. In particular the homology conclusion in Question 5 should not be substituted for the cohomology condition here.
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