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Let $BG$ is classifying space of $G$ topological group.

If $G$ is any compact group and $H$ is a closed subgroup of $G$, then the inclusion map $i:H\rightarrow G$ induces \begin{equation*} G/H\rightarrow BH\rightarrow BG \end{equation*} a fiber bundle?

If $G$ is any compact group and $H$ is a closed subgroup of $G$, then the inclusion map $i:H\rightarrow G$ induces
\begin{equation*} G/H\rightarrow BH\rightarrow BG \end{equation*} a fibration?

If $G$ is any compact group and $N$ is a closed normal subgroup of $G$, then the quotient map $\pi :G\rightarrow G/N$ induces
\begin{equation*} BN\rightarrow BG\rightarrow B\left( G/N\right) \end{equation*} a fiber bundle?

If $G$ is any compact group and $H$ is a closed normal subgroup of $G$, then the quotient map $\pi :G\rightarrow G/N$ induces \begin{equation*} BN\rightarrow BG\rightarrow B\left( G/N\right) \end{equation*} a fibration?

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    $\begingroup$ What is the difference between fiber bundle and fibration? $B$ is defined up to homotopy equivalence only, and, up to that, any map is a fibration. $\endgroup$
    – Alex Degtyarev
    Commented Apr 7, 2016 at 13:31
  • $\begingroup$ Do you have arbitrary compact topological groups? Do you have compact Lie groups? Which kind of fibration do you ask for? $\endgroup$
    – Helene Sigloch
    Commented Apr 7, 2016 at 13:55
  • $\begingroup$ This question shows I definitely need some more hypothesis for parts of this question. $\endgroup$
    – Omar Antolín-Camarena
    Commented Apr 7, 2016 at 18:20
  • $\begingroup$ $G$ is any compact topological group. not Lie group $\endgroup$
    – Mehmet Onat
    Commented Apr 8, 2016 at 8:16
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    $\begingroup$ Why is this on hold? What is unclear? Is it just because M. Onat wrote "fibration" instead of "fiber sequence" (it's true, as @AlexDegtyarev pointed out, that you might as well assume that $BH \to BG$ is a fibration, but then the question is whether the homotopy fiber is equivalent to $G/H$)? $\endgroup$
    – Omar Antolín-Camarena
    Commented Apr 8, 2016 at 17:10

1 Answer 1

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No, this is not true for Q1 and Q2.

To be specific we probably need to pick a particular model for $BG$, and there are several. For the argument I will give, the following property is sufficient: there exist isomorphisms $\pi_{i+1}(BG) \cong \pi_i(G)$ of homotopy groups.

Let $G = S^1$ and let $H$ be the subgroup of torsion elements (the roots of unity). The group $G$ is abelian so $H$ is normal, and $H$ is totally disconnected so $\pi_0(H) \cong H$ as sets. Any open set in $G$ intersects any coset $gH$ nontrivially, so the quotient space $G/H$ has the indiscrete topology and its homotopy groups vanish.

If the answer to any of the first two questions that you asked was affirmative, there would be an exact sequence in homotopy groups of the form $$ \cdots \to \pi_1(G/H) \to \pi_1(BH) \to \pi_1(BG) \to \cdots $$ and this would be an exact sequence $$ \cdots \to 0 \to H \to 0 \to \cdots $$ which is not possible.

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  • $\begingroup$ This nice example shows I definitely need more hypothesis for parts of this question. My guess is that $G \to G/H$ locally having a section is enough for Q2 here, do you think that's right? $\endgroup$
    – Omar Antolín-Camarena
    Commented Apr 7, 2016 at 18:20
  • $\begingroup$ Thanks Tyler Lawson. What do you think the last two question? $\endgroup$
    – Mehmet Onat
    Commented Apr 8, 2016 at 8:52
  • $\begingroup$ The Topology of Fiber Bundles Lecture Notes (Ralph L. Cohen) says that Proposition 2.15: Let $p:EG\rightarrow BG$ be a universal principal $G$ -bundle and $H<G$. Then there is a fiber bundle $BH\rightarrow BG$ with fiber the orbit space $G/H$. Proof: This bundle is given by $G/H\rightarrow EG\times _{G}G/H\rightarrow EG/G=BG$ together with the observation that $EG\times _{G}G/H=EG/H=BH.$ $\endgroup$
    – Mehmet Onat
    Commented Apr 8, 2016 at 8:57
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    $\begingroup$ @M.Onat I haven't looked in your reference but I'll venture a guess that $H$ is supposed to be closed $\endgroup$
    – Denis Nardin
    Commented Apr 8, 2016 at 17:06
  • $\begingroup$ I'm starting to think that $H$ closed may not be enough, @DenisNardin. Does that guarantee that $G \to G/H$ is a principal $H$-bundle, specifically, that it is locally trivial? I suspect that might be actually what you need for this. $\endgroup$
    – Omar Antolín-Camarena
    Commented Apr 8, 2016 at 17:14

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