Let $p\colon X\to B$ be a fibration. Let $G$ be a topological group acting continuously on $X$ and $B$, and assume that the map $p$ is $G$-equivariant. 


We can apply the Borel functor $EG\times\_G-$ for a contractible, free $G$-space $EG$. This gives a map $1\times\_G p\colon EG\times\_G X\to EG\times\_G B$.



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> Under what conditions on the group $G$ and the actions is the map $1\times\_G p$ guaranteed to be a fibration?
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References appreciated!



Does the Borel functor take equivariant fibrations to fibrations?
If $\pi:EG \to BG$ is a numerable $G$-principal bundle and if $p$ is a Hurewicz fibration, then the map $1 \times\_G p$ is a Hurewicz fibration.


Proof: Let $U \subset BG$ be open such that $EG \to BG$ is trivial over $U$. Observe that $\pi^{-1}(U) \times\_G B \cong U \times B$. Likewise for $X=E$ instead of $B$ and the map $1 \times\_G p:\pi^{-1}(U) \times\_G X \to \pi^{-1}(U) \times\_G B$ is the product of the identity on $U$ with $p:X \to B$. Hence $1 \times\_G p: \pi^{-1}(U) \times\_G X \to \pi^{-1}(U) \times\_G B$ is a Hurewicz fibration. Thus we can find a numerable covering of $EG \times\_G B$ such that the restriction of the map $1 \times\_G p$ to each of the covering maps is a Hurewicz fibration. 


To finish the proof, use the theorem that local Hurewicz fibrations are Hurewicz fibrations (see May, "A concise course in algebraic topology", p. 49 or tom Dieck "Algebraic topology", Theorem 13.4.1).


If you assume $p$ only to be a Serre fibration, you get that $1 \times\_G p$ is a Serre fibration with a similar argument.


