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I learnt this from Spanier and it is not very clear to me geometrically...

If I take a cohomology class in $H^n(X;G)$, is it possible for me to get an idea what exactly this map is in $[X;K(G,n)]$? For example, how can I possibly compute the homomorphism it induces $H^n(K(G,n);G)\to H^n(X;G)$?

Or conversely can I somehow think of a map that induces a given homomorphism on $H^n(K(G,n);G)\to H^n(X;G)$ as I wish? I know if I have $X=K(H,n)$ for the same $n$, it would be possible for me to transfer this info to the homotopy groups, but how about in general?

It seems I could refer to the simplicial construction of $K(G,n)$ (somebody recommended me to read May's Simplicial Objects in Algebraic Topology which does not feel to be a very easy reading for me) but I wish to hear some more thoughts if possible.

My example is $[\mathbb{T}^3;(\mathbb{C}P^\infty)^2]$ if this is helpful...

Thanks in advance!!

ah--
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  • @William Yes! Thank you! Edited! – ah-- Jul 10 '20 at 03:34
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    May's Simplicial Objects in Algebraic Topology is a somewhat dated book. – Connor Malin Jul 10 '20 at 03:37
  • @Connor I can somehow tell from the fact that there is no index at the end of the book lol... But would you elaborate in what sense you think it is dated? – ah-- Jul 10 '20 at 03:39
  • I believe it doesn't approach simplicial sets from the categorical view point, which is how everything now approaches it. – Connor Malin Jul 10 '20 at 03:40
  • I see and that's probably why the professor I asked recommended it to me because I asked for a more "geometric" way of looking at things... – ah-- Jul 10 '20 at 03:43

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If you give me a CW n-cocyle $\sigma$ on $X$, this is a map from the free abelian group on the cells to the group $G$. Now every map $f \in [X,K(G,n)]$ is homotopic to a map sending $X^{n-1}$ to a point, since $K(G,n)$ is (n-1)-connected. Thus, to give such a map $f$, up to homotopy it suffices to give a map $X/X^{n-1} \rightarrow K(G,n)$. The first thing we must do is describe what it does on the n-skeleton of this space. By elementary topology, the homotopy classes of maps from a wedge of spheres into $K(G,n)$ is a product over the spheres of $G$, which is the same as homming out from the free abelian group labeled by the cells.

We thus see that our cocycle $\sigma$ gives us a map on the n-skeleton of $X/X^{n-1}$. The fact that $\sigma$ is a cycle, one can check, is equivalent to the fact that this map extends to the (n+1)-skeleton. From there, one uses the fact that the higher homotopy groups vanish to extend the map to the rest of $X/X^{n-1}$, and there are no further obstructions.

Of course, one needs to check that these extensions are well defined up to homotopy.

So this is a formula for how to do it in general. If you want explicit maps, this will come down to calculating nullhomotopies of maps from spheres into your model of $K(G,n)$.

Connor Malin
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  • Thank you! Need some more time to think about it... But this reminds me of the way in which the universal element is constructed in Spanier. Any insights on what the "universal element" really is? (Sorry if the question doesn't make sense...) – ah-- Jul 10 '20 at 14:04
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    By the universal coefficient theorem plus Hurewicz, $H^n(K(G,n),G)=Hom(G,G)$, if we want to pick a good candidate for universal element it would be a good idea to pick the identity. In fact, by the Yoneda lemma (plus some other small stuff), up to an automorphism of G, the element we pick has to correspond to the identity element if the functor is to be representable. – Connor Malin Jul 10 '20 at 14:36
  • Sorry, I even feel shame to ask... Why does the cocycle $\sigma$ extend? – ah-- Jul 12 '20 at 06:26