0
$\begingroup$

Does there exist a fibration $K(\mathbb{Z}_4,1)\rightarrow K(\mathbb{Z}_2,1)$, evidently with fiber $K(\mathbb{Z}_2,1)$?

Improve this question
$\endgroup$
1
  • $\begingroup$ In homotopy theory, any map is a fibration. The one you are looking for is merely the one induced by the exact sequence of groups $1\to\Bbb Z_2\to\Bbb Z_4\to\Bbb Z_2\to0$. $\endgroup$
    – Alex Degtyarev
    Commented Mar 15, 2015 at 20:04

1 Answer 1

Reset to default
2
$\begingroup$

Every map can be replaced by a weakly equivalent fibration, using the path space (see this MSE question). Can you see how to use this to answer your question?

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks Tom. I was thinking about constructing explicit locally trivial bundle, I did not mention it in the question though. $\endgroup$
    – Samarkand
    Commented Mar 15, 2015 at 20:23
  • $\begingroup$ If you want a locally trivial fibre bundle, then you should say so in your question. A fibration (in the sense of Serre or Hurewicz) is a considerably weaker notion. In particular, path spaces are fibrations but (I think) not necessarily fibre bundles. $\endgroup$
    – ThiKu
    Commented Mar 16, 2015 at 2:16

Not the answer you're looking for? Browse other questions tagged .