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Jan 9, 2016 at 10:33 history closed Chris Gerig
José Figueroa-O'Farrill
András Bátkai
Wolfgang
Stefan Waldmann
Duplicate of "Dirty" proof that Eilenberg-MacLane spaces represent cohomology?
Jan 9, 2016 at 5:55 comment added Ryan Budney Yes, of course there's intuition. It's in the proof! The proof is about as elementary as you could hope for a theorem of this type. Understand the proof and you will understand much more than this theorem. The technique of the proof is very much the "initial spirit" of obstruction theory. A wonderfully productive subject that gave birth to many productive careers.
Jan 8, 2016 at 22:19 comment added Dylan Wilson You can take, as a model for $K(\mathbb{Z},n)$, the free abelian group on $S^n$ where the basepoint acts as the identity. Then, given an $n$-cocycle $\phi$ you define a map on $X$ by sending the $(n-1)$-skeleton to the basepoint and then sending an $n$-cell $\sigma$ to the sphere by modding out by the boundary, then multiply the result by $\phi(\sigma)$. This gets you to the $n$-skeleton... to extend further you use that $\phi$ was a cocycle.
Jan 8, 2016 at 22:09 review Close votes
Jan 9, 2016 at 10:33
Jan 8, 2016 at 21:46 comment added Chris Gerig Also here: mathoverflow.net/questions/5518/…
Jan 8, 2016 at 21:37 comment added Chris Gerig This was answered here: mathoverflow.net/questions/2890/… To repeat: $H^n(K(G,n);G)=Hom(\pi_nK(G,n),G)=Hom(G,G)$ and so there is a distinguished element $u\in H^n(K(G,n);G)$ corresponding to the identity $1:G\to G$. The bijection is given by pull-back, $f\mapsto f^*u$. For your simple example ($G =\mathbb{Z}$, $n = 1$), take $c \in H^1(X)$ to map the 1-skeleton of $X$ to $S^1$, where an edge $e$ will make $c(e)$ loops around $S^1$.
Jan 8, 2016 at 20:28 review First posts
Jan 8, 2016 at 20:49
Jan 8, 2016 at 20:22 history asked Krishna CC BY-SA 3.0