# $G/H$ with $H$ contractible a fibre bundle?

Suppose $G$ is a topological group and $H\leq G$ a normal/closed subgroup of $G$. If $H$ is contractible, does the quotient map $p: G\rightarrow G/H$ form a fibre bundle?

Is there a more general condition on $G$ or $H$ that guarantees that the map $p: G\rightarrow G/H$ is a fibre bundle? References for both of these questions will be warmly received

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By fiber bundle you mean locally trivial fiber bundle? –  Mariano Suárez-Alvarez Oct 18 '11 at 6:30
@Mariano Yes I mean it is locally trivial. Could you provide me with an example of one that is not? I guess I always thought this was one of the defining characteristics of a fiber bundle. –  mjones Oct 18 '11 at 8:20
@mjones - perhaps this MO question will help? –  Juan S Oct 18 '11 at 8:36

I'm not sure about the $H$ contractible part.
For the general question, have a look at these notes of Peter May's. Proposition 3.7 states that if $H$ has local cross-sections in $G$ then $p$ will be a principal $H$ bundle. According to Wikipedia this is also in Steenrod's book on fibre bundles. Wiki also has the following
The most general conditions under which the quotient map will admit local cross-sections are not known, although if $G$ is a Lie group and $H$ a closed subgroup (and thus a Lie subgroup by Cartan's theorem), then the quotient map is a fiber bundle. One example of this is the Hopf fibration, $S^3 \to S^2$ which is a fiber bundle over the sphere $S^2$ whose total space is $S^3$. From the perspective of Lie groups, $S^3$ can be identified with the special unitary group $SU(2)$. The abelian subgroup of diagonal matrices is isomorphic to the circle group $U(1)$, and the quotient $SU(2)/U(1)$ is diffeomorphic to the sphere.
More generally, if $G$ is any topological group and $H$ a closed subgroup which also happens to be a Lie group, then G → G/H is a fiber bundle.