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Let $p\colon X\to B$ be a fibration. Let $G$ be a topological group acting continuously on $X$ and $B$, and assume that the map $p$ is $G$-equivariant.

We can apply the Borel functor $EG\times_G-$ for a contractible, free $G$-space $EG$. This gives a map $1\times_G p\colon EG\times_G X\to EG\times_G B$.

Under what conditions on the group $G$ and the actions is the map $1\times_G p$ guaranteed to be a fibration?

References appreciated!

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Probably a stupid question, but what are the natural maps $B\to G$ and $E\to G$ that we're pulling back? –  Harry Gindi Jan 16 '11 at 15:17
@Harry: in the present context, $EG \times_G B:= (EG \times B)/G$ denotes the Borel construction, not the pullback. I admit that this notational ambiguity is slightly annyoing, but it is completely standard. I prefer to use the symbol $E(G;B)$. –  Johannes Ebert Jan 16 '11 at 17:10
Harry, it's actually not a pullback, but rather a quotient by the action of G. While I agree the notation can be confusing, it's also fairly common. I personally prefer to put the G above the $\times$ so that it doesn't look like a pullback, but I'm probably in the minority. –  Mike Skirvin Jan 16 '11 at 17:11
But the notation $E(G;B)=EG \times_G B$, unfortunately, only applies to the universal bundle, while the construction can be done for each principal $G$-bundle and $G$-space. –  Johannes Ebert Jan 16 '11 at 17:11

2 Answers 2

up vote 7 down vote accepted

If $\pi:EG \to BG$ is a numerable $G$-principal bundle and if $p$ is a Hurewicz fibration, then the map $1 \times_G p$ is a Hurewicz fibration.

Proof: Let $U \subset BG$ be open such that $EG \to BG$ is trivial over $U$. Observe that $\pi^{-1}(U) \times_G B \cong U \times B$. Likewise for $X=E$ instead of $B$ and the map $1 \times_G p:\pi^{-1}(U) \times_G X \to \pi^{-1}(U) \times_G B$ is the product of the identity on $U$ with $p:X \to B$. Hence $1 \times_G p: \pi^{-1}(U) \times_G X \to \pi^{-1}(U) \times_G B$ is a Hurewicz fibration. Thus we can find a numerable covering of $EG \times_G B$ such that the restriction of the map $1 \times_G p$ to each of the covering maps is a Hurewicz fibration.

To finish the proof, use the theorem that local Hurewicz fibrations are Hurewicz fibrations (see May, "A concise course in algebraic topology", p. 49 or tom Dieck "Algebraic topology", Theorem 13.4.1).

If you assume $p$ only to be a Serre fibration, you get that $1 \times_G p$ is a Serre fibration with a similar argument.

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Thanks, that's great! –  Mark Grant Jan 17 '11 at 13:09
By the way, is this proof written down somewhere I can refer to? –  Mark Grant Jan 20 '11 at 12:42
I have no idea where to find this written down. –  Johannes Ebert Jan 20 '11 at 13:54

I believe what you are asking is true whenever $X\to B$, considered as an unequivariant map, is a Serre fibration.

First some definitions:

Call a map of $G$-spaces $E \to B$ a $G$-Serre fibration if and only if for all subgroups $H\subset G$, the map of fixed points $E^H \to B^H$ is a Serre fibration. In particular, $EG \times E \to EG \times B$ is a $G$-Serre fibration if and only if it is a Serre fibration of unequivariant spaces. It is known this notion of fibration arises from a model structure on $G$-spaces, in which a map $X\to Y$ is a $G$-weak equivalence iff each map of fixed point sets $X^H \to Y^H$ is a weak homotopy equivalence. A map $X\to Y$ is a $G$-cofibration iff $Y$ is obtained from $X$ by attaching cells of the form $D^n \times (G/H)$ where $H$-varies through subgroups and the attaching maps are $G$-maps, or more generally if the pair $(Y,X)$ is a retract of a relative $G$-cell complex.

I do not know a reference for the above, but I am confident it's in the literature. (Added Later: see the comment below for two references.)

Now the argument:

Suppose that $A\to U$ is an acyclic cofibration in the Serre model structure on spaces. Without loss in generality, we can assume that $U$ is obtained from $A$ by cell attachments. Suppose we are given a lifting problem:

$A \to X \times_G EG $

$\downarrow\qquad \qquad \downarrow$

$U \to B\times_G EG$

We need to find a map $U \to X\times_G EG$ making the diagram commute.

Here's how: pull back the above to an equivariant lifting problem

$\tilde A \to X \times EG $

$\downarrow\qquad \qquad \downarrow$

$\tilde U \to B\times EG$

where $\tilde A$ for example is given by the pullback of $A \to BG \leftarrow EG$ (the map $A\to BG$ is the composite $A\to X\times_G EG \to BG$). It is relatively straightforward to check that the inclusion $\tilde A\to \tilde U$ is an acyclic $G$-cofibration, where the cells that are being attached are of the form $D^n \times G$, i.e., they're free.

It follows from the model category structure on $G$-spaces that there's an equivariant lift $\tilde U \to X\times EG$ making the diagram commute. Now take orbits to get the a lift $$ U \to X\times_G EG $$ solving the original lifting problem.

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Here are two references for the model structure: 1. Dwyer, W.G., Kan, D.M.: Singular functors and realization functors. Nederl. Akad. Wetensch. Indag. Math. 46 147–153 (1984) (nd.edu/~wgd/Dvi/SingularAndRealization.pdf) and also Ch. VI §5 of the following: 2. May, J.P., et. al.: Equivariant Homotopy and Cohomology Theory. CBMS Regional Conference Series in Mathematics 91, 1996. –  John Klein Jan 17 '11 at 2:39
I wish I could accept both these answers. I chose Johannes' because I'm interested in the case $G=S^1$, but it's nice to know it holds in full generality. –  Mark Grant Jan 17 '11 at 13:12

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