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This is a substantial revision of my original post. It shows that if we replace the "equivalence" Tyler is asking for by a "retract" then the answer is yes.

1. Given a CW space $Y$, we can take $G(Y) =$ the topological monoid of homotopy automorphisms of $Y$. The Borel construction $$EG(Y) \times_{G(Y)} Y \to BG(Y)$$ is then a quasifibration. Let $U \to BG(Y)$ be the effect of converting it into a fibration.

2. Let $G$ be a topological group with a chosen homotopy equivalence $$BG\simeq BG(Y).$$ For example, we can do what Tyler does, or we can simply take $\Omega BG(Y)$, where this means the realized Kan loop of the total singular complex of $BG(Y)$.

3. Let $EG \to BG$ be a universal $G$-principal bundle, and set $$Z \quad := \quad \text{pullback}(EG \to BG \simeq BG(Y) \leftarrow U)$$ Then $Z \subset EG \times U$ inherits a $G$-action and its underlying homotopy type is that of $Y$. Then the Borel construction $$EG\times_G Z \to BG$$ is a fiber bundle which is weak fiber homotopy equivalent to $U \to BG(Y)$.

4. Step 3 implies that $BG(Y)$ is a retract up to homotopy of $B\text{homeo}(Z)$. This will imply that $G(Y)$ is a homotopy retract of $\text{homeo}(Z)$ in the $A_\infty$ sense, with $Z \simeq Y$.

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This is a comment because it takes space)substantial revision of my original post. Someone once told me Peter May showed It shows that a fibration is always fiber homotopy equivalent to some fiber bundle (where the fiber can be pretty large). I don't know how this is proved, and I am not sure if I am even remembering we replace the statement correctly. But here goes...

Assuming it "equivalence" Tyler is correct, asking for by a "retract" then to every fibration $E \to B$ with fiber $X$ which the answer is classified by yes.

• Given a map CW space $B \to BG(X)$, where Y$, we can take$G(X)$is G(Y) =$ the topological monoid of homotopy automorphisms , there is supposed to be a homotopy equivalence $Y \to X$ and a factorization up to homotopy of the classifying map:B \to B\text{aut}(Y) \to BG(Y) \simeq BG(X) .$$So if we consider the universal (quasi-)fibration Y. The Borel constructionF \to EG(XEG(Y) \times_{G(X)} F times_{G(Y)} Y \to BG(XBG(Y)$$with classifying map the identity $BG(X) \to BG(X)$, it will factorize through $B\text{aut}(Y)$ for some choice of $Y$. This would show at least that $BG(X)$ is then a retract of quasifibration. Let $B\text{aut}(Y)$.

Note added: the result I attributed U \to Peter May is more-or-less contained in BG(Y)$be the effect of converting it into a paper by Casson and Gottliebfibration.The idea is this: any finite CW • Let$X$is homotopy equivalent to G$ be a codimension zeroopen submanifold $M \subset \Bbb R^n$ for topological group with a chosen homotopy equivalence $n$ large (this is given by thickening)$BG\simeq BG(Y). They show if$n$is large that [B,B\text{diff}(M)] \to [B,BG(M)] = [B,BG(X)]$$is surjectiveFor example, where B is finite dimensional CW and n depends on \dim X and \dim B.This shows that any finite skeleton of BG(X) is a retract of B\text{diff}(M \times \Bbb R^j) for suitable choice of j. Since B\text{diff}(M) \to BG(M) factors through B\text{homeo}(M), we get the same statement for homeomorphisms. Finallycan do whatTyler does, or we can simply take j \Omega BG(Y), where this means the realized Kan loop the total singular complex of BG(Y). • Let EG \to \infty. This will show thatBG(X) is BG be a retract of universal B\text{homeo}^{\text{st}}(M), where G-principal bundle, and set Z \text{homeo}^{\text{st}}(M) quad := \lim_j \quad \text{homeo}(M\times text{pullback}(EG \Bbb R^jto BG \simeq BG(Y) .\leftarrow U)By the way, this last group maps to Then \text{homeo}(M Z \subset EG \times Q) where U inherits a Q G-action and its underlying homotopy type is the Hilbert cube. So we get that BG(X) is a retract of B\text{homeo}(M \times Q). Yet another note: A possibly related result due to Chapman: If X is 1-connected CW, then Y. Then the map\text{homeo}(X\times Q) EG\times_G Z \to G(X \times Q) \simeq G(X)is 1-connected, in particular, any self-homotopy equivalence f: X\to X a fiber bundle which is such thatweak fiber homotopy equivalent to f\times 1_Q: X \times Q U \to X\times Q BG(Y). • Step 3 implies that BG(Y) is homotopic a retract up to homotopy of B\text{homeo}(Z).This will imply that G(Y) is a homeomorphismhomotopy retract of \text{homeo}(Z) in the A_\infty sense, with Z \simeq Y. • 8 added 4 characters in body The following might lead to an answer of your question (I am posting it as an answer instead of a comment because it takes space). Someone once told me Peter May showed that a fibration is always fiber homotopy equivalent to some fiber bundle (where the fiber can be pretty large). I don't know how this is proved, and I am not sure if I am even remembering the statement correctly. But here goes... Assuming it is correct, then to every fibration E \to B with fiber X which is classified by a map B \to BG(X), where G(X) is the monoid of homotopy automorphisms, there is supposed to be a homotopy equivalence Y \to X and a factorization up to homotopy of the classifying map:$$ B \to B\text{aut}(Y) \to BG(Y) \simeq BG(X) . $$So if we consider the universal (quasi-)fibration$$ F \to EG(X) \times_{G(X)} F \to BG(X) $$with classifying map the identity BG(X) \to BG(X), it will factorize through B\text{aut}(Y) for some choice of Y. This would show at least that BG(X) is a retract of B\text{aut}(Y). Note added: the result of I attributed to Peter May I quoted is more-or-less contained in a paper by Casson and Gottlieb. The idea is this: any finite CW X is homotopy equivalent to a codimension zero open submanifold M \subset \Bbb R^n for n large (this is given by thickening). They show if n is large that$$ [B,B\text{diff}(M)] \to [B,BG(M)] = [B,BG(X)] $$is surjective, where B is finite dimensional CW and n depends on \dim X and \dim B. This shows that any finite skeleton of BG(X) is a retract of B\text{diff}(M \times \Bbb R^j) for suitable choice of j. Since B\text{diff}(M) \to BG(M) factors through B\text{homeo}(M), we get the same statement for homeomorphisms. Finally, we can take j \to \infty. This will show that BG(X) is a retract of B\text{homeo}^{\text{st}}(M), where$$ \text{homeo}^{\text{st}}(M) = \lim_j \quad \text{homeo}(M\times \Bbb R^j) . $$By the way, this last group maps to \text{homeo}(M \times Q) where Q is the Hilbert cube. So we get that BG(X) is a retract of B\text{homeo}(M \times Q). Yet another note: A possibly related result due to Chapman: If X is 1-connected CW, then the map$$ \text{homeo}(X\times Q) \to G(X \times Q) \simeq G(X)$$is$1$-connected, in particular, any self-homotopy equivalence$f: X\to X$is such that$f\times 1_Q: X \times Q \to X\times Q\$ is homotopic to a homeomorphism.

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