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How does the inclusion $\mathbb Z\rightarrow \mathbb Q$ induce a fibration $K(\mathbb Z,n)\rightarrow K(\mathbb Q,n)$ with fibre $\Omega K(\mathbb Q/\mathbb Z,n)$?

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This really isn't a great question, since it is not at all clear where your difficulty lies. One can define a functor $K(-, n)$ (as in my answer), and that's really all there is to it. (I had trouble deciding whether to answer at all, and whether this question would be better suited for math.stackexchange.com. It's definitely not a research-level question; see the faq.) –  Todd Trimble Jun 8 '11 at 9:48
    
This is a very simple exercise. –  Fernando Muro Jun 8 '11 at 9:57
    
I have now deleted my answer, since the question has been changed. You have to choose your models correctly to get this fiber in a point-set topology sense, but it isn't hard. –  Todd Trimble Jun 8 '11 at 10:36
    
It's still the same very simple exercise I used to solve as an undergraduate student. –  Fernando Muro Jun 8 '11 at 11:11
    
@Fernando Muro : very simple?? :) thanks anyway fernando. –  palio Jun 8 '11 at 11:19
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up vote 3 down vote accepted

Probably the most functorial approach is to use the Dold-Kan equivalence $$F:\{\text{chain complexes}\} \to \{\text{simplicial abelian groups}\}. $$ Let $A_{\ast}$ denote the chain complex with just $\mathbb{Q}/\mathbb{Z}$ in dimension $n-1$, let $B_{\ast}$ be the one with a surjective differential from $\mathbb{Q}$ in dimension $n$ to $\mathbb{Q}/\mathbb{Z}$ in dimension $n-1$, and let $C_{\ast}$ be the one with just a $\mathbb{Q}$ in dimension $n$. There is an evident short exact sequence (and therefore fibration) $A_{\ast}\to B_{\ast}\to C_{\ast}$, which gives a fibration $|FA_{\ast}|\to |FB_{\ast}|\to |FC_{\ast}|$ of topological abelian groups. Here $|FA_{\ast}|$ and $|FC_{\ast}|$ are $K(\mathbb{Q}/\mathbb{Z},n-1)$ and $K(\mathbb{Q},n)$ essentially by definition, and it is easy to produce a weak equivalence from the corresponding model for $K(\mathbb{Z},n)$ to $|FB_{\ast}|$.

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