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 


*

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

*$f$ 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 ?  
 A: 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.
A: 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.
