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Let $K(m, \mathbb{Z})$ be the Eilenberg-MacLane space. I've read that $BK(m, \mathbb{Z}) = K(m+1, \mathbb{Z})$, but I'm trying to understand this.

I'm familiar with the bar construction $BG$ for a group $G$: roughly, we start with a point, attach 1-cells labelled by elements of $G$, and then we glue in a $n$-cells with edges $g_1, g_2, \ldots, g_n$ and $g_1g_2\cdots g_n$. (Please correct me if that's not quite right). But what does $BK(m, \mathbb{Z})$ mean? I know that by cup product we have a map $K(m, \mathbb{Z}) \times K(n, \mathbb{Z}) \to K(m+n, \mathbb{Z})$ but I don't know if this is relevant.

Najib Idrissi
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usr0192
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  • Hint: what's $K(1,\mathbb{Z})$ ? What does the cup product do to spaces? – Benjamin Gadoua Jun 30 '16 at 01:30
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    There is a topological group model of $K(G,m)$ when $G$ is abelian, so you take the classifying space of this. Then use the fact that for a topological group, $G \simeq \Omega BG$. –  Jun 30 '16 at 01:44
  • @BenjaminGadoua $K(1, \mathbb{Z}) = S^1$. I don't really understand the cup product - given two cochains, it is the cochain where you apply the first to the front face, the second to the back face, and then take the product of the two numbers. But I don't see what you are trying to suggest. – usr0192 Jun 30 '16 at 02:57
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    @MikeMiller Thanks. The $G \simeq \Omega BG$ tells me that $BK(m, \mathbb{Z}) = K(m+1, \mathbb{Z})$. But I didn't know that $K(G, m)$ has an abelian group model for $G$ abelian...is this easy to see? I mean, what is the map $K(m, G) \times K(m, G) \to K(m, G)$? Given two degree $m$ cohomology classes, I need to construct a degree $m$ class. Perhaps take their sum? – usr0192 Jun 30 '16 at 03:05
  • @user90219 Have a look at Peter May's answer here: http://math.stackexchange.com/questions/45556/group-structure-on-eilenberg-maclane-spaces – iwriteonbananas Jun 30 '16 at 13:22

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