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?


1 Answer 1


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 $B(F,\operatorname{haut}(F),\operatorname{haut}(F)) \rightarrow E' \rightarrow X$.

  • $\begingroup$ Thanks! But I have some question: What is the $\Omega_{Moore}$? And where could I find the detail of correspondence between fibrations and the homotopy class $[X, B\ haut(F)]$ $\endgroup$
    – Cloudifold
    Commented Jun 19, 2022 at 8:37
  • $\begingroup$ I accidentally called these $\Omega_{Moore}$ when I should have called it $\Omega_{Kan}$. What I mean by $\Omega_{Kan}$ is a certain model of the loop space, constructed using simplicial sets, which is actually a topological group. Search "Kan loop group" for a construction. $\endgroup$ Commented Jun 19, 2022 at 15:58
  • $\begingroup$ By the clutching construction, we know that fibrations over $\Sigma X$ correspond to maps $[X,\operatorname{haut}(F)]$. Now consider $[X,\Omega B \operatorname{haut}(F)]$. By adjunction, this is $[\Sigma X, B \operatorname{haut}(F)]$, but it is also $[X,\operatorname{haut}(F)]$ since $\Omega$ and $B$ are inverse for grouplike monoids. Hence, $[\Sigma X, B\operatorname{haut}(F)]$ classifies fibrations over $\Sigma X$. So at least on the class of suspensions, it classifies fibrations. $\endgroup$ Commented Jun 19, 2022 at 16:06
  • $\begingroup$ For the general case, it is theorem 1.3 in "THE CLASSIFYING SPACES FOR SURGERY AND COBORDISM OF MANIFOLDS" where it is attributed to Stasheff. $\endgroup$ Commented Jun 19, 2022 at 16:09
  • $\begingroup$ By the way, here is a paper which proves the result of your question jstor.org/stable/2046409?seq=1 $\endgroup$ Commented Jun 19, 2022 at 16:13

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