# Fiber bundle and fibration of classifying space [closed]

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?

• 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. – Alex Degtyarev Apr 7 '16 at 13:31
• Do you have arbitrary compact topological groups? Do you have compact Lie groups? Which kind of fibration do you ask for? – Helene Sigloch Apr 7 '16 at 13:55
• This question shows I definitely need some more hypothesis for parts of this question. – Omar Antolín-Camarena Apr 7 '16 at 18:20
• $G$ is any compact topological group. not Lie group – Mehmet Onat Apr 8 '16 at 8:16
• 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 $G/H$)? – Omar Antolín-Camarena Apr 8 '16 at 17:10

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.

• 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? – Omar Antolín-Camarena Apr 7 '16 at 18:20
• Thanks Tyler Lawson. What do you think the last two question? – Mehmet Onat Apr 8 '16 at 8:52
• 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 $EG\times _{G}G/H=EG/H=BH.$ – Mehmet Onat Apr 8 '16 at 8:57
• @M.Onat I haven't looked in your reference but I'll venture a guess that $H$ is supposed to be closed – Denis Nardin Apr 8 '16 at 17:06
• 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. – Omar Antolín-Camarena Apr 8 '16 at 17:14