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.


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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