A homogeneous space that's not a fibre bundle Let $G$ be a locally-compact group and $H$ a closed subgroup.
Let $X=G/H$ and let $\pi:G\to X$ be the projection.
We say that the projection $\pi$ is a fibre bundle, if for every point $x\in X$ there exists an open neighborhood $U\subset X$ and a continuous map $s:U\to G$ such that $p\circ s=Id_U$.
Paul Mostert proved in 1956, that every $X$ is a fibre bundle under some condition of finite-dimensionality.
I would like to see an example of a pair $(G,H)$ such that the projection fails to be a fibre bundle.
 A: Let $G=\prod_{j=1}^\infty{\mathbb T}$, where ${\mathbb T}=\{z\in{\mathbb C}:|z|=1\}$ is the circle group and $G$ is equipped with the product topology.
Let $H=\prod_{j=1}^\infty\{\pm 1\}$.
Then $H$ is a closed subgroup, but $p:G\to G/H$ is not a fibre bundle.
To see this assume there is a continuous section $s:U\to G$ for some open neighborhood $U$ of $eH$ in $G/H$.
The space $G/H=\prod_{j=1}^\infty{\mathbb T}/\{\pm 1\}$ carries the product topology.
Therefore, $U$ contains an open subset of the form $V=\prod_{j=1}^{N-1}V_j\times\prod_{j=N}^\infty {\mathbb T}/\{\pm 1\}$, where each $V_j$ is an open unit-neighborhood in ${\mathbb T}$.
Let $i:{\mathbb T}/\{\pm 1\}\to \prod_{j=1}^\infty{\mathbb T}/\{\pm 1\}$ be the injection at the $N$-th coordinate, i.e., $i(z)=(1,\dots,1,z,1,\dots)$ with the $z$ in the $N$-th place.
Let $p:\prod_{j=1}^\infty{\mathbb T}\to{\mathbb T}$ the projection onto the $N$-th coordinate.
Then the map $\sigma: {\mathbb T}/\{\pm 1\}\to{\mathbb T}$, given by
$$
\sigma=p\circ s\circ i
$$
is a continuous section to the projection ${\mathbb T}\to{\mathbb T}/\{\pm 1\}$, but such a continuous section does not exist. Contradiction.
