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This question already has an answer here:

Hatcher's AT Theorem 4.57 is used in both the algebraic topology construction of Seifert surfaces, and the (similarly flavored) proof that given a compact 3-manifold (with or without boundary), we can associate a properly embedded surface $S$ to any $[a] \in H_2(M, \partial M; \mathbb{Z})$ (this is used to define the Thurston Norm).

Theorem 4.57 states (direct quote): there are natural bijections $T: \langle X, K(G,n) \rangle \to H^n(X; G)$, for all CW-complexes $X$ and all $n > 0$, with $G$ any abelian group. Such a $T$ has the form $T([f]) = f^*(\alpha)$ for a certain distinguished class $\alpha \in H^n(K(G,n); G)$ (Hatcher, Algebraic Topology, Page 393). (Notation: $\langle X, K(G,n) \rangle$ is the set of basepoint-preserving homotopy classes of maps from $X$ to a $K(G,n)$).

The proof (to me, at least) is somewhat complicated, and is constructed by creating a cohomology theory, and uses some basic formalisms (that is, once you get through the 8 pages of algebraic topology, the proof of 4.57 falls out in 1 paragraph).

My question: the $T$ above is a bijection; suppose we assume $X$ is something nice, like a smooth manifold, and $K(G,n)$ is also relatively simple, perhaps $S^1$, which is a $K(\mathbb{Z}, 1)$ -- is there some topological or geometric way to see how a function $f \in \langle X, K(G,n)\rangle$ is built in association with $[a] \in H^n(X,\mathbb{Z})$?

A more general (and fluffy) question: again assuming $X, K(G,n)$ are "nice", as above, is there some intuition for why this theorem holds? (I understand that this is perhaps a somewhat unreasonable question).

Thank you!

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marked as duplicate by Chris Gerig, José Figueroa-O'Farrill, András Bátkai, Wolfgang, Stefan Waldmann Jan 9 '16 at 10:33

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  • $\begingroup$ 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 $S^1$. $\endgroup$ – Chris Gerig Jan 8 '16 at 21:37
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    $\begingroup$ Also here: mathoverflow.net/questions/5518/… $\endgroup$ – Chris Gerig Jan 8 '16 at 21:46
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    $\begingroup$ 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. $\endgroup$ – Dylan Wilson Jan 8 '16 at 22:19
  • $\begingroup$ 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. $\endgroup$ – Ryan Budney Jan 9 '16 at 5:55