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This came up in the question about Eilenberg-MacLane spaces. Given the definition of
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If you form the smash product X = K(A,p) n K(B,q) of two Eilenberg-MacLane spaces (I am using "n" for the smash product symbol here), then the resulting space is (p+q-1)-connected, and the first non-trivial homotopy group in dimension p+q is A \otimes B. To see this "geometrically", I would model the EM spaces as CW-complexes, whose first non-basepoint cells are in dimensions p and q respectively. Then in the smash product X, the bottom dimensional cells are products of the bottom dimensional cells of the EM spaces; this gives the connectivity result, and by looking at the attaching maps of the (p+q+1)-dimensional cells in X, you can compute \pi_{p+q}. Now you can use obstruction theory to produce a map X --> K(A\otimes B, p+q). |
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One perspective I like is that homology is abelianization. Specifically, given a based space X you can construct a new space AG(X), the (reduced) free abelian group on X, whose points are finite formal sums Σ nx xof points of X with integer coefficients, subject to the relation that the basepoint is sent to 0. (The topology takes some describing, but it is roughly a quotient topology from two copies of the infinite symmetric product.) Instead of integer coefficients I could take coefficients in some abelian group M, but let's stick to this for now. For X a CW-complex, the homotopy groups of AG(X) are actually the reduced integral homology groups of X (this is the Dold-Thom theorem). You might believe this because you can show that the functor X -> π*(AG(X)) is a homology theory satisfying the Eilenberg-Steenrod axioms, because it converts quotient sequences X -> Y -> Y/X of spaces into exact sequences AG(X) -> AG(Y) -> AG(Y/X) of topological groups, which are (almost) fibration sequences. Under this perspective, you actually have a construction of K(Z,n); it is AG(Sn), the free abelian group on the n-sphere. The map described in previous answers from K(Z,n) ^ K(Z,m) to K(Z,n+m), giving you cup products, is then the map AG(Sn) ^ AG(Sm) -> AG(Sn ^ Sm) = AG(Sn+m) given by (Σ nx x) ^ (Σ my y) -> Σ (nx my) (x ^ y) which looks exactly like a tensor product map. |
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A very specific example: for G=Z and n=m=1, it is the map S^1 \times S^1 \to CP^\infty which maps S^1 \times S^1 to the 2-skeleton S^2 of CP^\infty by a degree 1 map. Here's a general but less geometric way of understanding it. K(G,n) can be constructed as the realization of the simplicial abelian group associated to the chain complex with Z in degree n and 0 elsewhere under the Dold-Kan correspondence. By a form of Eilenberg-Zilber, the levelwise tensor product of two simplicial abelian groups is, up to natural weak equivalence, the same as the ordinary tensor product of chain complexes. But the ordinary tensor product takes K(G,n) and K(G,m) to K(G \otimes G,m+n), and there is a natural map from K(G,n) \times K(G,m) to the levelwise tensor product given levelwise by the natural map from a product to a tensor product. Composing all of this together with a ring multiplication map G \otimes G \to G giving K(G \otimes G,m+n)\to K(G,m+n) should give the cup product. With some unraveling that I don't have time to do right now but I invite someone else to figure out, you should be able to turn this into a reasonably explicit map on the simplicial abelian group level coming from the Alexander-Whitney map. |
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There is a geometric (space level) way of realizing Eric's answer. In fact, this is the subject of section 1 in a paper by Ravenel and Wilson (MathSciNet). They used the iterated simplcial bar construction B^nG as a model for K(G,n) so that, when G is a ring, the ring multiplication G\times G -> G induces G\times BG -> BG by fixing the first factor and then BG\times BG -> B(BG)=K(G,2) by fixing the second factor, and so on. An explicit description is found in page 700 in their paper, where they prove the resulting maps give rise to the cup product pairing (Theorem 1.7). |
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