Is every fibration fiber homotopy equivalent to a fiber bundle? A fibration $p : E \to B $ over a contractible base B is fiber homotopy equivalent to a product fibration $B \times F \to B$. (Corollary 4.63. Hatcher's Algebraic Topology)
A locally trivial bundle(or fiber bundle) with fiber $F$ is a map $p' : E \to B$
such that forall $b \in B$, exists neighbourhood $U$ of $b$
with $p'^{-1}(U) \cong U \times F$, and $U \times F \xrightarrow{pr_1} U$ commutes with $p'^{-1}(U) \xrightarrow{p'} U$.
I want to prove that any fibration $p : E \to B$ is fiber homotopy equivalent to a locally trivial bundle.
(Assume $E,B$ are compactly generated Hausdorff spaces, and $B$ have a numerable open contractible cover)
But I meet some obstruction in patching homotopy equivalences on $E_U \to U$ to a global homotopy equivalence.(where $U$ is in the open contractible  cover of $B$)
Is that true or have a counter example?
 A: Every fibration $p:F \rightarrow E \rightarrow B$ is equivalent to a fiber bundle, but not necessarily with any representative of the homotopy type of $F$. For example, spherical fibrations are not equivalent to sphere bundles since the J-homomorphism is not surjective.
Recall that given a topological monoid $M$, a right $M$ module $R$, and a left $M$ module $L$, we can form $B(R,M,L)$, the bar construction of $M$ and $L$.
Now remember there is a correspondence between fibrations over $X$ with fiber $F$ and $[X,B\operatorname{haut}(F)]$, the classifying space of the monoid of homotopy automorphisms of $F$. There is an equivalence of monoids between $\operatorname{haut}(F)$ and $\Omega_{\operatorname{Kan}} B \operatorname{haut}(F)$, the important fact being that the latter is a strict topological group.
So we deduce that fibrations over $X$ with fiber $F$ are in correspondence with $[X,B\Omega_{\operatorname{Kan}} B \operatorname{haut}(F)]$. The latter classifies $\Omega_{\operatorname{Kan}} B \operatorname{haut}(F)$ bundles over $X$. At this point, we have shown there is a correspondence with fibrations and principal bundles. Now we want to extract a fiber bundle out of the latter with the homotopy type of the fiber being $F$.
Consider $B(F,\operatorname{haut}(F),\operatorname{haut}(F))$. Sweeping a little under the rug, the equivalence $\operatorname{haut}(F) \simeq \Omega_{\operatorname{Kan}} B \operatorname{haut}(F)$ endows the bar construction with a right $\Omega_{\operatorname{Kan}} B \operatorname{haut}(F)$ action because $B \operatorname{haut}(F)$ is a bimodule over itself.  By replacing the fiber, prinicpal $\Omega_{\operatorname{Kan}} B \operatorname{haut}(F)$ bundles are in correspondence with $B(F,\operatorname{haut}(F),\operatorname{haut}(F))$ bundles with structure group $\Omega_{\operatorname{Kan}} B \operatorname{haut}(F)$. Now classically $F \simeq B(F,\operatorname{haut}(F),\operatorname{haut}(F))$ and this respects the right module structures, akin to the fact that $M \otimes_R R \cong M$ as right $R$-modules.
To summarize: given a fibration $F \rightarrow E \rightarrow X$ we constructed a fiberwise equivalent fiber bundle $B(F,\operatorname{haut}(F),\operatorname{haut}(F)) \rightarrow E' \rightarrow X$.
