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edited Jan 25 2011 at 1:52 |
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.
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. 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 $BG(Y)$. 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 $U \to BG(Y)$. 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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edited Jan 24 2011 at 19:29 |
The following might lead to an answer of your question (I am posting it as an answer instead of 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 that$BG(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$.
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edited Jan 23 2011 at 23:01 |
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)
$$
is $1$-connected, in particular, any self-homotopy equivalence $f: X\to X$ is such that
$f\times 1_Q: X \times Q \to X\times Q$ is homotopic to a homeomorphism.
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edited Jan 23 2011 at 22:48 |
The following might lead to an answer to of your question (I am posting it as an answer instead of a comment because it takes space). Someone once told me Peter May showed that a fibration is always fiber homotopy equivalent to some fiber bundle (where the fiber can be pretty large). I don't know how this is proved, and I am not sure if I am even remembering the statement correctly. But here goes...
Assuming it is correct, then to every fibration $E \to B$ with fiber $X$ which is classified by a map $B \to BG(X)$, where $G(X)$ is the monoid of homotopy automorphisms, there is supposed to be a homotopy equivalence $Y \to X$ and a factorization up to homotopy of the classifying map:
$$
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)
$$
with classifying map the identity $BG(X) \to BG(X)$, it will factorize through $B\text{aut}(Y)$ for some choice of $Y$. This would show at least that $BG(X)$ is a retract of $B\text{aut}(Y)$.
Note added: the result of Peter May I quoted is more-or-less contained in a paper by Casson and Gottlieb. The idea is this: any finite CW $X$ is homotopy equivalent to a codimension zero
open submanifold $M \subset \Bbb R^n$ for $n$ large (this is given by thickening).
They show if $n$ is large that
$$
[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)
$$
is $1$-connected, in particular, any self-homotopy equivalence $f: X\to X$ is such that
$f\times 1_Q: X \times Q \to X\times Q$ is homotopic to a homeomorphism.
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edited Jan 23 2011 at 17:52 |
The following might lead to answer to your question (I am posting it as an answer instead of a comment because it takes space). Someone once told me Peter May showed that a fibration is always fiber homotopy equivalent to some fiber bundle (where the fiber can be pretty large). I don't know how this is proved, and I am not sure if I am even remembering the statement correctly. But here goes...
Assuming it is correct, then to every fibration $E \to B$ with fiber $X$ which is classified by a map $B \to BG(X)$, where $G(X)$ is the monoid of homotopy automorphisms, there is supposed to be a homotopy equivalence $Y \to X$ and a factorization up to homotopy of the classifying map:
$$
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)
$$
with classifying map the identity $BG(X) \to BG(X)$, it will factorize through $B\text{aut}(Y)$ for some choice of $Y$. This would show at least that $BG(X)$ is a retract of $B\text{aut}(Y)$.
Note added: the result of Peter May I quoted is more-or-less contained in a paper by Casson and Gottlieb. The idea is this: any finite CW $X$ is homotopy equivalent to a codimension zero
open submanifold $M \subset \Bbb R^n$ for $n$ large (this is given by thickening).
They show if $n$ is large that
$$
[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) .
$$
By the way, this last group maps to $\text{homeo}(M \times Q)$ where $M$ Q$ is the Hilbert cube. So we get that $BG(X)$ is a retract of $B\text{homeo}(M \times Q)$.
Yet another note: A possibly related result due to Chapman:
If $X$ is $1$-connected CW, then the map
$$
\text{homeo}(X\times Q) \to G(X \times Q) \simeq G(X)
$$
is $1$-connected, in particular, any self-homotopy equivalence $f: X\to X$ is such that
$f\times 1_Q: X \times Q \to X\times Q$ is homotopic to a homeomorphism.
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edited Jan 23 2011 at 8:37 |
The following might lead to answer to your question (I am posting it as an answer instead of a comment because it takes space). Someone once told me Peter May showed that a fibration is always fiber homotopy equivalent to some fiber bundle (where the fiber can be pretty large). I don't know how this is proved, and I am not sure if I am even remembering the statement correctly. But here goes...
Assuming it is correct, then to every fibration $E \to B$ with fiber $X$ which is classified by a map $B \to BG(X)$, where $G(X)$ is the monoid of homotopy automorphisms, there is supposed to be a homotopy equivalence $Y \to X$ and a factorization up to homotopy of the classifying map:
$$
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)
$$
with classifying map the identity $BG(X) \to BG(X)$, it will factorize through $B\text{aut}(Y)$ for some choice of $Y$. This would show at least that $BG(X)$ is a retract of $B\text{aut}(Y)$.
Note added: the result of Peter May I quoted is more-or-less contained in a paper by Casson and Gottlieb. The idea is this: any finite CW $X$ is homotopy equivalent to a codimension zero
open submanifold $M \subset \Bbb R^n$ for $n$ large (this is given by thickening).
They show if $n$ is large that
$$
[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) .
$$
By the way, this last group maps to $\text{homeo}(M \times Q)$ where $M$ is the Hilbert cube. So we get that $BG(X)$ is a retract of $B\text{homeo}(M \times Q)$.
Yet another note: A possibly related result due to Chapman:
If $X$ is $1$-connected CW, then the map
$$
\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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edited Jan 23 2011 at 8:05 |
Note added: the result of Peter May I quoted is also more-or-less contained in a paper by Casson and Gottlieb. They give a geometric argument along the following lines. Suppose that $E\to B$ The idea is a fibration with fiber $X$ a this: any finite CW complex. Then $X$ can be thickened is homotopy equivalent to a compact codimension zeroopen submanifold $M \subset \Bbb R^n$ for some $n$ depending on $\dim X$.The derivative map \text{diff}(M)\to F(M,O_n)large (this is then defined. Let $\text{diff'}(M)$ be its homotopy fibergiven by thickening). These are the diffeomorphisms They show if $f$ n$ is large that $M$ equipped with a path from $df\: M [B,B\text{diff}(M)] \to O_n$ to the identity. Using Smale-Hirsch theory[B,BG(M)] = [B,BG(X)]is surjective, it's possible to show that the map where $\text{diff'}(M)\to G(M)$B$ is highly connected: the range of dimensions depends on finite dimensional CW and $n$ and depends on $\dim X$ , but it tends to infinity with and $n$ (where we stabilize \dim B$.This shows that any finite skeleton of $M$ by BG(X)$ is a retract of $M B\text{diff}(M \times \Bbb R$ to increase R^j)$ for suitable choice of $n$). Consequently, if j$. Since $n$ is taken very large and B\text{diff}(M) \to BG(M)$ factors through $B$ is finite dimensionalB\text{homeo}(M)$, we get the same statement for homeomorphisms. Finally, we can find take $j \to \infty$. This will show that$BG(X)$ is a homotopy factorizationretract of $B\text{homeo}^{\text{st}}(M)$, where $B $\to B\text{diff'}(M text{homeo}^{\text{st}}(M) = \times lim_j \Bbb R^n) quad \to B\text{diff}(M\times text{homeo}(M\times \Bbb R^nR^j) \.By the way, this last group maps to BG(M)$\text{homeo}(M \times Q)$ where $This shows M$ is the Hilbert cube. So we get that any fibration with a finite CW fiber over a finite dimensional base space lifts to $BG(X)$ is a fiber bundleretract of $B\text{homeo}(M \times Q)$.
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edited Jan 23 2011 at 7:40 |
Note added: the result of Peter May I quoted is also contained in a paper by Casson and Gottlieb. They give a geometric argument along the following lines. Suppose that $E\to B$ is a fibration with fiber $X$ a finite CW complex. Then $X$ can be thickened to a compact codimension zero open submanifold $M \subset \Bbb R^n$ for some $n$ depending on $\dim X$.The derivative map \text{diff}(M)\to F(M,O_n)is then defined. Let $\text{diff'}(M)$ be its homotopy fiber. These are the diffeomorphismsof $f$ $M$ equipped with a path from $df\: M \to O_n$ to the identity. Using Smale-Hirsch theory, it's possible to show that the map $\text{diff'}(M)\to G(M)$is highly connected: the range of dimensions depends on $n$ and $\dim X$, but it tends to infinity with $n$ (where we stabilize $M$ by $M \times \Bbb R$ to increase $n$). Consequently, if $n$ is taken very large and $B$ is finite dimensional, we can find a homotopy factorizationB \to B\text{diff'}(M \times \Bbb R^n) \to B\text{diff}(M\times \Bbb R^n) \to BG(M)This shows that any fibration with a finite CW fiber over a finite dimensional base space lifts to a fiber bundle.
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edited Jan 23 2011 at 7:02 |
The following might lead to answer to 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 lifts up to fiber homotopy equivalence equivalent to a 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)
$$
with classifying map the identity $BG(X) \to BG(X)$, it will factorize through $B\text{aut}(Y)$ for some choice of $Y$. This would show at least that $BG(X)$ is a retract of $B\text{aut}(Y)$.
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answered Jan 23 2011 at 6:51 |
The following might lead to answer to your question (I am posting it as an answer instead of a comment because it takes space). Someone once told me Peter May showed that a fibration always lifts up to fiber homotopy equivalence to a fiber bundle (where the fiber can be pretty large). I don't know how this is proved, and I am not sure if I am even remembering the statement correctly. But here goes...
Assuming it is correct, then to every fibration $E \to B$ with fiber $X$ which is classified by a map $B \to BG(X)$, where $G(X)$ is the monoid of homotopy automorphisms, there is supposed to be a homotopy equivalence $Y \to X$ and a factorization up to homotopy of the classifying map:
$$
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 $B\text{aut}(Y)$.
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