1
$\begingroup$

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

$\endgroup$
7
  • 4
    $\begingroup$ 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.) $\endgroup$
    – Todd Trimble
    Jun 8, 2011 at 9:48
  • $\begingroup$ This is a very simple exercise. $\endgroup$ Jun 8, 2011 at 9:57
  • $\begingroup$ 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. $\endgroup$
    – Todd Trimble
    Jun 8, 2011 at 10:36
  • $\begingroup$ It's still the same very simple exercise I used to solve as an undergraduate student. $\endgroup$ Jun 8, 2011 at 11:11
  • 1
    $\begingroup$ @Fernando Muro : very simple?? :) thanks anyway fernando. $\endgroup$
    – palio
    Jun 8, 2011 at 11:19

1 Answer 1

3
$\begingroup$

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 $|FB_{\ast}|$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.