Reduced Vs unreduced cohomology in the parametrized setting.

Question

Can someone explain the relationship between reduced and unreduced parametrized homology theories in the parametrized setting à la May-Sigurdsson with maps to a reference space $B$?. Is it just a copy of the "classical cohomology theory" on the reference space $B$. How do you go between reduced and unreduced theories using the base change functors?

Answer 1

If $(X,p)$ is a space over $B$, its unreduced cohomology is the same as the reduced cohomology of $(X,p)_+ = (X\sqcup B,p,\sigma)$, $\sigma$ being the section taking $B$ to the disjoint copy of $B$.

But I suspect you're interested in the relationship between the reduced and unreduced cohomologies of a given ex-space $(X,p,\sigma)$. We can assume that it's well-sectioned. The section $\sigma$ gives a cofibration sequence $(B,\mathrm{id})_+ \to (X,p)_+ \to (X,p,\sigma)$, and projection gives a retraction $(X,p)_+\to (B,\mathrm{id})_+$. Therefore, the unreduced cohomology of $X$ splits as the direct sum of its reduced cohomology and the unreduced cohomology of $B$.

Which leads to the question of what the unreduced cohomology of $B$ is, as a space over itself. Consider the "trivial" case, where the cohomology theory is pulled back from a nonparametrized theory along the projection $B\to *$ to a point. (That is, its representing spectrum is given as such.) Then there is a change-of-base adjunction that implies that the parametrized cohomology of $B$ is just the nonparametrized cohomology of $B$. In general, though, it's going to be something like twisted cohomology.