Taking homotopy fixed points preserves fibrations I'm reading a paper where they have an appendix about homotopy fixed point sets of a $G$-space, and at some point they claim that if $f:X\to Y$ is a $G$-map that is an ordinary (non-equivariant) fibration between $G$-spaces, then $f^{hG} : X^{hG}\to Y^{hG}$ is still a fibration; and the proof is "for general reasons".
Note that the model used for $X^{hG}$ is $\mathrm{Map}^G(EG,X)$ (this statement is probably not model-independent, because if I understand correctly, any homotopy equivalent space could play the role of the homotopy fixed points)
What can the authors mean by "general reasons" ? I can see easily that $\mathrm{Map}(EG,X)\to \mathrm{Map}(EG,Y)$ is a fibration, maybe what I'm missing is that it's an equivariant fibration ? 
If it is, then I can conclude, because of course when $f: E\to B$ is an equivariant fibration (I haven't seen that terminology but I assume it would mean that it has the $G$-homotopy lifting property against $G$-spaces, right ? - I'll use that interpretation, if I'm wrong please tell me so), then $f^G :E^G\to B^G$ is a fibration; and $\mathrm{Map}^G(EG,X)$ is just $\mathrm{Map}(EG,X)^G$; but I don't see why it would be an equivariant fibration, when $X\to Y$ is only an ordinary fibration...
 A: Here's my second shot at this question.  Hopefully it turns out better.  
We want to show that $X^{hG} \to Y^{hG}$ is a fibration, using the model $X^{hG} = \operatorname{Map}^G(EG,X)$.  
Now, by definition, $X^{hG} \to Y^{hG}$ is a fibration iff it has the rlp against acyclic cofibrations of the form $A \to A \times I$.  Here we assume that $A$ is a CW complex.  
Since $X^{hG} = \operatorname{Map}^G(EG,X)$ can be thought of as the (normal) $G$-fixed points of the conjugation action on $\operatorname{Map}(EG,X)$, we see that $X^{hG} \to Y^{hG}$ is a fibration if $\operatorname{Map}(EG,X) \to \operatorname{Map}(EG,Y)$ has the rlp in the category of $G$-spaces and $G$-maps against acyclic cofibrations $A \to A \times I$, where $A$ and $A \times I$ are given the trivial $G$-action.  
By the exponential adjunction, this happens iff $X \to Y$ (again in the category of $G$-spaces) has the rlp against $A \times EG \to A \times I \times EG$.  Note that since $G$ acts freely on $EG$, the isotropy of these spaces is contained in the trivial group $e$.  
We will show that $X \to Y$ has the lifting property by induction on the cells of $A$.  Suppose we have constructed the desired lift over a skeleton of $A$ in $A \times I \times EG$.  Let $\sigma$ be a cell of $A$ over which we want a lift.  Then by assumption that $X \to Y$ is a (non-equivariant) fibration, there is a lift over $X^e \to Y^e$, which extends uniquely to the $G$-orbit of $\sigma$ in $A \times I \times EG$ - this orbit is free, so there are no problems.  Then rinse and repeat with the other cells. 
There probably is a way to say this last part cleanly using a notion of lifting properties with respect to isotropies, but I haven't figured this out.
