10
$\begingroup$

It is well known that the functor of cohomology is representable. More precisely, given $n\ge1$ and abelian group $G$, we have $H^n(X;G)\simeq[X,K(G,n)]$. (Here we probably need some ``nice'' assumptions, e.g. we work in (homotopy) category of $CW$-complexes, or some other.)

My question: Is there some similar theorem for cohomology with local coefficients?

Apparently, we need to consider the subcategory of $\mathrm{hTop}$ where morphisms preserve the given local system. However, it is more interesting to directly find an analogue of the Eilenberg-MacLane space $K(G,n)$ for local coefficients $G$ (which action contains all automorphisms of $G$?) and prove the analogue of the Brown's theorem using the obstruction theory technique (instead of category-theoretical one).


I found a construction of $K_\pi(G,n)$ space in this mo question. If we set $\pi=\mathrm{Aut}(G)$ and take $\psi:\pi_1(X)\to\pi$, is it correct that $$ \{f\in[X,K_\pi(G,n)]:\pi_1(f)=\psi\} \simeq H^n(X;f^*\pi_n(K_\pi(G,n)))? $$ If so, can we prove this using obstructions?

$\endgroup$
2
  • 2
    $\begingroup$ I'm not sure if the formula you write on the bottom makes sense. What is $f$ on the right hand side? $\endgroup$ Oct 30, 2022 at 17:19
  • 1
    $\begingroup$ @R.vanDobbendeBruyn here for $f$ we can take every map which induces the homomorphism $\psi$ on $\pi_1$ --- for all such maps corresponding pullbacks of $G$ as a local system are canonically isomorphic. $\endgroup$ Oct 30, 2022 at 18:51

2 Answers 2

12
$\begingroup$

Yes. This is covered in the wonderful book Lecture Notes in Algebraic Topology by Davis and Kirk. Specifically, in Theorem 5.12 they define the relevant category the OP alludes to, and in Theorem 5.13 they prove that cohomology with local coefficients satisfies the axioms of a homology theory, e.g., excision. All you need is those axioms. With them in hand, you can follow the argument explained in Mark Grant's answer to Is every homology theory given by a spectrum?, which references Hatcher's book and Switzer's, to see that the functor is representable.

$\endgroup$
2
  • $\begingroup$ Your link to the question Is every homology theory given by a spectrum? seemed to be referencing @MarkGrant's answer, so I edited accordingly while this was on the front page. I hope that was all right. $\endgroup$
    – LSpice
    Mar 26 at 14:07
  • 1
    $\begingroup$ @LSpice Sure, thanks! $\endgroup$ Mar 26 at 15:18
2
$\begingroup$

Floris van Doorn's Ph.D. thesis On the Formalization of Higher Inductive Types and Synthetic Homotopy Theory gives a very nice account of parametrized cohomology (which generalizes cohomology with local coefficients) in the language of HoTT. Basically, cohomology theories (ordinary and extraordinary) study the mapping spectrum $\operatorname{Hom}(X,E)$ where $E$ is a spectrum. On the other hand, parametrized cohomology theories study the spectrum of sections $\Gamma(X,E)$, where now $E$ is a fibration by spectra over $X$, which we can think of as a family of spectra parametrized by $X$. In the case where the fibres are Eilenberg–MacLane spectra, we get cohomology with local coefficients.

To see how this connects with more classical approaches to local coefficients, suppose that we have a locally constant sheaf of abelian groups, which corresponds to a covering space whose fibres are abelian groups by the usual etale space construction. Now we apply the Eilenberg–MacLane space construction fibrewise to get a fibration by Eilenberg–MacLane spectra.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.