# Fibration of principal bundles

Let $$G$$ be a topological group, let $$f:X\rightarrow Z$$ be a $$G$$-equivariant map of (left) $$G$$-spaces such that

1. $$X\rightarrow X/G$$ and $$Z\rightarrow Z/G$$ are principal $$G$$-bundles.

2. $$f$$ is a fibration.

Let $$\rho: G\rightarrow H$$ be a morphism of topological groups. Is the induced map $$H\times_{G}X \rightarrow H\times_{G}Z$$ a fibration ? where the (right) action of $$G$$ on $$H$$ is induced by $$\rho$$

Edit: I was trying the following proof, but I think it is incomplete may be someone could help.

First: $$f$$ induces a continuous map of topological spaces $$\hat{f}: X/G\rightarrow Z/G$$. The pullback of the map $$Z\rightarrow Z/G$$ along $$\hat{f}: X/G\rightarrow Z/G$$ is exactly (up to isomorphism) the map $$X\rightarrow X/G$$.

Second: The pullback of the map $$H\times_{G}Z\rightarrow Z/G$$ along $$\hat{f}$$ is exactly (up to isomorphism) the map $$H\times_{G}X\rightarrow X/G$$.

In the first and second item we use the fact that we have $$G$$-principal bundles and $$H$$-principal bundles.

If we can proof that $$\hat{f}$$ is a fibration then we are done! So my question would be answered if $$\hat{f}$$ is a fibration. If $$f$$ is a fibration does it follow that $$\hat{f}$$ is a fibration ?

• What exactly do you mean by "map" of $G$-principal bundles? If it is the usual definition (equivariant and commutes with the projection maps) then it must be an isomorphism... – John Greenwood Feb 3 at 5:36
• @JohnGreenwood I have edited my question, I hope it is clear. – GSM Feb 3 at 7:45
• Is $f$ $G$-equivariant? – Ben McKay Feb 3 at 8:37
• @BenMcKay yes f is G-equivariant. – GSM Feb 3 at 8:42
• Can we assume Z/G is paracompact? – John Greenwood Feb 3 at 18:20

The previous answer was getting a bit too complicated. Locally $$f$$ looks like $$O_i \times G \rightarrow U_i \times G$$ and this is a fibration since it is the restriction of the original fibration $$f$$. Now $$\hat f$$ locally looks like $$O_i \rightarrow U_i$$, and this is a fibration since it is a retract of a fibration. Thus locally $$\hat f$$ is a fibration, and so, thanks to Dold, $$\hat f$$ is a fibration.

• Ah, It was your first claim I was having trouble convincing myself of! But "Dold" still requires numerability right? – John Greenwood Feb 6 at 4:10

I think the answer is "yes" if $$Z/G$$ is paracompact and there is a cover by contractible neighborhoods over which $$p_{Z}: Z\rightarrow Z/G$$ is trivial.

By paracompactness such a cover admits a numerable refinement $$\{U_{i}\}$$, and by a standard theorem it suffices to check that $$\hat{f}$$ is a fibration over each $$U_{i}$$.

Now $$p_{Z}^{-1}(U_{i})\simeq U_{i}\times G$$ by assumption, and $$U_{i}$$ is contractible. Let $$F$$ be the fiber of $$f$$. Let $$g:G\rightarrow Z$$ be the orbit of some point in $$p_{Z}^{-1}(U_{i})$$. Then $$f^{-1}(p_{Z}^{-1}(U_{i}))\simeq U_{i}\times g^{*}X$$. Note that the second factor is an $$F$$-fibration over $$G$$.

Now suppose we're given a homotopy $$H:Y\times I\rightarrow U_{i}$$ and a lift $$h_{0}:Y\times\{0\}$$ to $$\hat{f}^{-1}(U_{i})$$. Clearly $$h_{0}$$ lifts to $$p_{Z}^{-1}(U_{i})$$, and by the above it also lifts to $$f^{-1}(p_{Z}^{-1}(U_{i})$$. Since the composite $$X\rightarrow Z\rightarrow Z/G$$ is a fibration, we get a lift of the homotopy $$H$$ to $$X$$. Composing with the projection to $$X/G$$ and commutativity of the obvious square involving $$X,Z,X/G$$ and $$Z/G$$ provides us with a lift of $$H$$ to $$X/G$$ that lands in $$\hat{f}^{-1}(U_{i})$$ as desired.

• Thanks! Where do you use that the spaces Z/G has a cover by contractible neighborhoods? is it a crucial assumption ? – GSM Feb 3 at 18:59
• I use it to split off a factor of U_i in the last equality of the second paragraph, I'm not sure how crucial it is to the general problem though! – John Greenwood Feb 3 at 19:08