Assume that $H$ is a closed subgroup of a Lie Group $G$ and let $EG \rightarrow BG$ the universal bundle.

It is well known the fact that there is an induced map $BH \rightarrow BG$ with homotopy fiber $G/H$. The proof is also an standard argument: $BH \cong EG \times_G G/H \rightarrow BG$ can be seen as the "associate" fiber bundle of the universal bundle $EG \rightarrow BG$.

It is also a well known fact that the classifying space construction is functorial, so the inclusion $i: H \rightarrow G$ induces a map $Bi: BH \rightarrow BG$.

My question is, why is this map $Bi$ a fibration with fiber $G/H$ and does it coincide (up to homotopy) with the first construction?

In general, the map $Bi$ arises as the classifying map of the principal $G$-bundle $EG \times_H G \rightarrow BH$, so my idea is to show that the pullback of the map $EG \times_G G/H \rightarrow BG$ is actually $EG \times_H G$ but I had no success.


1 Answer 1


Firstly, this is not the correct context to talk about strict fibrations, since the classifying space functor is only defined up to homotopy equivalence. Whilst I believe that there are constructions that make the induced map $BH\rightarrow BG$ into a Serre fibration for a closed subgroup $H\leq G$, these constructions are by no means unique. Rather, one should talk about homotopy fibrations in this context, which are simply the result of turning an arbitrary map into a fibration (say by pulling back from a path space fibration). The point is that this procedure correctly identifies the homotopy type of the homotopy fibre of $Bi$ without asserting the existence of any homotopy lifting property or bundle structure.

Now, with that out the way, let us address the question. Let $G$ be a suitable topological group. Assuming it is compact Lie is certainly more than adequate. Then, as you say, there is a functorially associated classifying space $BG$ satisfying a certain universal property. However, more is true, and in many cases it is useful to take this extra structure into account. Not just the classifying space $BG$ is functorial, but there is a functorially associated $G$-principal fibration

$$G\rightarrow EG\xrightarrow{\pi_G}BG$$

where $EG$ is a contractible free $G$-space. Let us write $\mathcal{E}G$ for this principal bundle. Then a homomorphism $\varphi:G\rightarrow G'$ will induce a morphism of fibre bundles


The induced map on fibres will be homotopic to $\varphi$, and the induced map of base spaces is a map $B\varphi:BG\rightarrow BG'$ that classifies $\varphi$.

Note that the sense of this functorality is homotopical, and although the bundle is defined strictly, the spaces $EG$ and $BG$ and maps $B\varphi$ are only defined up to a suitable notion of homotopy.

Now let $i:H\hookrightarrow G$ be a subgroup. Then there is an induced map $\mathcal{E}i:\mathcal{E}H\rightarrow\mathcal{E}G$ as above, and the map of total spaces $Ei:EH\rightarrow EG$ is an $H$-equivariant map which is a non-equivariant homotopy equivalence, since $EH$, $EG$ are both non-equivariantly contractible.

Moreover, since it is a subgroup, $H$ acts freely on $EG$ through this map and induces a map

$$Bi':BH=(EH)/H\rightarrow (EG)/H$$

of quotient spaces, which is easily seen to be a homotopy equivalence using the universal properties. On the other hand, since $H\leq G$, there is an induced map of orbit spaces

$$Bi'':(EG)/H\rightarrow (EG)/G=BG$$

and we easily see that

$$Bi=Bi''\circ Bi'.$$

Now we write

$$Bi'': EG/H\cong (EG\times_GG)/H\cong EG\times_G(G/H)\xrightarrow{\pi_G} BG$$

and use this to identify the homotopy fibre of $Bi''$ as $G/H$. Since $Bi=Bi''\circ Bi'$, and $Bi$ is a homotopy equivalence, it follows that the homotopy fibre of $Bi$ is also $G/H$.

  • $\begingroup$ Thanks @Tyrone. It is a very detailed answer, so much clear now. $\endgroup$
    – C. Zhihao
    Sep 15, 2018 at 19:04

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