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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 ?

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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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