E is a bundle space over B relative to $p$ iff there is a local cross-section of B in E .

Question

A map $p : E \rightarrow B $ is said to have the bundle property, if there exists a space D such that, for each $b \in B$, there is an open neighborhood $U$ of $b$ in $B$ together with a homeomorphism

$ \phi_{U} : U \times D \rightarrow p^{-1}(U)$ of $ U \times D$ onto $p^{-1}(U)$ satisfying the condition:

$ p \phi_{U}(u,d)=u$ for $(u \in U, d \in D)$

In this case, the space E is called a bundle space over the base space B relative to the projection $p : E \rightarrow B$.

Suppose E is a topological group and If F is a closed subgroup of E and $B=E\F$, whose elements are left cosets of F in E that is B is a homogeneous space , then

Now,I don't know how to proceed:

"E is a bundle space over B relative to p iff there is a local cross-section of B in E ; by this we mean a cross-section $x : V \rightarrow E$ defined on an open neighbourhood V of the point $ b_0 \in F$ in B"

One side is clear, that is if E is a bundle space over B relative to p then from the definition of bundle space we have a local cross section namely $\phi :U \times {b_0} \rightarrow p^{-1}(U)$ and $p\phi(u,b_0)=u$ hence we are done for this case. Now, how to proceed the other way? Thanks for any help!

Answer 1

Your statement (Edit: before you corrected it) is false:

Consider the first projection $$p:E=\mathbb R^2\setminus \{(0,0)\}\to B=\mathbb R:(x,y)\mapsto x$$ The map $p$ has local and even global cross sections, like for example$$s:B\to E:x\mapsto (x,17)$$ But $p$ is certainly not a bundle since the fibre $p^{-1}(0)=\mathbb R^*$ of $0\in B$ is not homeomorphic to the other fibres $p^{-1}(x)=\mathbb R\; (x\neq0)$ of $p$.
Thus no space $D$ can be found to satisfy the conditions required for $p:E\to B$ to be a bundle.

Remark
It is however true that a principal bundle is trivial if and only it has a global cross section.
However in this theorem you have to assume you have a bundle to start with, and even a very special sort of bundle.