G$ a fibration if $G$ acts freely on $X$?

Asked Jul 21 '14 at 11:09

Active Jul 21 '14 at 11:14

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Suppose that $G$ acts freely on $X$, and let $EG$ be a contractible space on which $G$ acts freely. According to many references, the projection $(X\times EG)/G\to X/G$ is a fibration.

However, I can't find a proof for this, nor construct one myself.

edited Jul 21 '14 at 11:14

asked Jul 21 '14 at 11:09

Bashar Saleh

Hint: the fiber is $EG/G$. – Vladimir Jul 21 '14 at 11:57
The fiber is $EG$? Can I determine if a map is a fibration by looking at the fibers? – Bashar Saleh Jul 21 '14 at 12:01
The fiber is $EG/G$, not $EG$. 2) A more profound hint: $(X\times EG)/G=(X/G)\times (EG/G)$. 3) Just in case: do you know the definition of a fibration?

Vladimir

Jul 21 '14 at 12:03

I'm sorry, I forgot to say that I equipped $X\times EG$ with the diagonal $G$-action, i.e. the action given by $g.(x,e)= (gx,ge)$, so $(X\times EG)/G$ is not $X/G\times EG/G$. 3) According to Hatcher: A fibration is a map $p: E\to B$ having the homotopy lifting property w.r.t. all spaces $X$.

– Bashar Saleh Jul 21 '14 at 12:11

Why is $(X\times EG)/G\to X/G$ a fibration if $G$ acts freely on $X$?

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