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In Spanier's book " Algebraic topology" a fiber bundle is defined as follows: A fiber bundle $\xi=(E,B,F,p)$ consists of a total space $E$, a base space $B$ and a fiber $F$ and a bundle projection $p:E\rightarrow B$ such that there is exists an open covering $\{U\}$ of $B$ and for each $U\in \{U\}$ a homeomorphism $\phi_U: U\times F\rightarrow p^{-1}(U)$ such that the composite $p\circ \phi_U: U\times F\rightarrow U$ is projection map.

Question: If a compact lie group G (e.g circle group) acts freely on a paracompact space X then does $(X,X/G,G,\pi)$ (where $X/G$ is orbit space and $\pi$ is orbit map) forms a fiber bundle.

If $\pi$ is covering projection then it seems the above is true but what I am not able to verify this is in general case.

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1 Answer 1

The answer is yes if you assume paracompact spaces are Hausdorff, and no if not.

This is because in Hausdorff spaces, paracompactness implies complete regularity, and it is a theorem of Gleason that if compact Lie group $G$ acts on a completely regular space $X$, then every orbit $Gx$ admits a tube neighbourhood. This implies that a free action is a principal $G$-bundle in this case. See also Bredon's "Introduction to compact transformation groups", Theorem II.5.8.

On the other hand, its easy to cook up counter-examples for non-Hausdorff paracompact spaces. In fact, just take your favourite free $G$-space $X$, and give $X$ the indiscrete topology.

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