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 ?

  • 1
    $\begingroup$ 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... $\endgroup$ – John Greenwood Feb 3 at 5:36
  • $\begingroup$ @JohnGreenwood I have edited my question, I hope it is clear. $\endgroup$ – GSM Feb 3 at 7:45
  • $\begingroup$ Is $f$ $G$-equivariant? $\endgroup$ – Ben McKay Feb 3 at 8:37
  • $\begingroup$ @BenMcKay yes f is G-equivariant. $\endgroup$ – GSM Feb 3 at 8:42
  • $\begingroup$ Can we assume Z/G is paracompact? $\endgroup$ – 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.

  • $\begingroup$ Ah, It was your first claim I was having trouble convincing myself of! But "Dold" still requires numerability right? $\endgroup$ – 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.

  • $\begingroup$ Thanks! Where do you use that the spaces Z/G has a cover by contractible neighborhoods? is it a crucial assumption ? $\endgroup$ – GSM Feb 3 at 18:59
  • $\begingroup$ 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! $\endgroup$ – John Greenwood Feb 3 at 19:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.