I know that Eilenberg-Maclane spaces $K(G,n)$ are classifying spaces for cohomology meaning that $H^n(X,G) \cong [X,K(G,n)]$ but is there a way to retrieve the cohomology product from this description?
Ring structure of cohomology from homotopy classes of maps into Eilenberg-Maclane spaces
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algebraic-topology
1 Answers
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Yes.
There is a map $m:K(G, m)\times K(G, n)\to K(G, n+m)$, which gives the map map $H^n(X; G)\times H^m(X; G)=[X, K(G, n)]\times [X, K(G, m)]=[X, K(G, n+m)]\to [X, K(G, n+m)]=H^{n+m}(X; G)$, which gives the ring structure. In fact, we can recover the map by applying the Yoneda embedding to the cup-product, though it is important to note that this method only determines the multiplication, and thus the communitivity diagrams up to homotopy.
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0Great! Can you explain how this map m is constructed or point to somewhere where I could read it? – 2017-02-08
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1The map is constructed using the Yoneda embedding. $m:[X, K(G, n)\times K(G, m)]=H^m(X; G)\times H^n(X, G)\to H^{n+m}(X; G)=[X, K(G, n+m)]$, so letting $X=K(G, n)\times K(G, m)$, we get $[X, X]\to [X, K(G, n+m)]$, so the identity map is sent to our desired multiplication map. You can give an explicit construction though. Namely, you can provide an explicit construction of $K(G, m)$ inductively using the Bar-resolution and then you can write down the map almost trivially (this has the benefit of functionally giving the map strictly, rather than up to homotopy). – 2017-02-09