fibre of a fibration is homotopy equivalent to its homotopy fibre
$begingroup$
Can someone give me a hint on proving that the fibre of a fibration $f: Y to X$ is homotopy equivalent to its homotopy fibre $Y times_X X^I$?
general-topology homotopy-theory
asked May 10 '11 at 11:27
user5262user5262
1,002713
$endgroup$
add a comment |
$begingroup$
Can someone give me a hint on proving that the fibre of a fibration $f: Y to X$ is homotopy equivalent to its homotopy fibre $Y times_X X^I$?
general-topology homotopy-theory
asked May 10 '11 at 11:27
user5262user5262
1,002713
$endgroup$
1
$begingroup$
Not a hint, but a couple of arguments and references can be found in this MO-thread.
$endgroup$
– t.b.
May 10 '11 at 11:55
$begingroup$
Thanks, I already found this. But I don't want the complete argument.
$endgroup$
– user5262
May 10 '11 at 12:33
$begingroup$
The first step would be to construct a continuous function from one to the other. There appears to be a good choice from the fibre to the homotopy fibre.
$endgroup$
– Ryan Budney
May 10 '11 at 17:38
add a comment |
$begingroup$
Can someone give me a hint on proving that the fibre of a fibration $f: Y to X$ is homotopy equivalent to its homotopy fibre $Y times_X X^I$?
general-topology homotopy-theory
asked May 10 '11 at 11:27
user5262user5262
1,002713
$endgroup$
Can someone give me a hint on proving that the fibre of a fibration $f: Y to X$ is homotopy equivalent to its homotopy fibre $Y times_X X^I$?
general-topology homotopy-theory
general-topology homotopy-theory
asked May 10 '11 at 11:27
user5262user5262
1,002713
asked May 10 '11 at 11:27
user5262user5262
1,002713
asked May 10 '11 at 11:27
user5262user5262
1,002713
asked May 10 '11 at 11:27
user5262user5262
1,002713
asked May 10 '11 at 11:27
user5262user5262
1,002713
1,002713
1
$begingroup$
Not a hint, but a couple of arguments and references can be found in this MO-thread.
$endgroup$
– t.b.
May 10 '11 at 11:55
$begingroup$
Thanks, I already found this. But I don't want the complete argument.
$endgroup$
– user5262
May 10 '11 at 12:33
$begingroup$
The first step would be to construct a continuous function from one to the other. There appears to be a good choice from the fibre to the homotopy fibre.
$endgroup$
– Ryan Budney
May 10 '11 at 17:38
add a comment |
1
$begingroup$
Not a hint, but a couple of arguments and references can be found in this MO-thread.
$endgroup$
– t.b.
May 10 '11 at 11:55
$begingroup$
Thanks, I already found this. But I don't want the complete argument.
$endgroup$
– user5262
May 10 '11 at 12:33
$begingroup$
The first step would be to construct a continuous function from one to the other. There appears to be a good choice from the fibre to the homotopy fibre.
$endgroup$
– Ryan Budney
May 10 '11 at 17:38
1
1
$begingroup$
Not a hint, but a couple of arguments and references can be found in this MO-thread.
$endgroup$
– t.b.
May 10 '11 at 11:55
$begingroup$
Not a hint, but a couple of arguments and references can be found in this MO-thread.
$endgroup$
– t.b.
May 10 '11 at 11:55
$begingroup$
Thanks, I already found this. But I don't want the complete argument.
$endgroup$
– user5262
May 10 '11 at 12:33
$begingroup$
Thanks, I already found this. But I don't want the complete argument.
$endgroup$
– user5262
May 10 '11 at 12:33
$begingroup$
The first step would be to construct a continuous function from one to the other. There appears to be a good choice from the fibre to the homotopy fibre.
$endgroup$
– Ryan Budney
May 10 '11 at 17:38
$begingroup$
The first step would be to construct a continuous function from one to the other. There appears to be a good choice from the fibre to the homotopy fibre.
$endgroup$
– Ryan Budney
May 10 '11 at 17:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is a consequence of the main theorem in
R. Brown and P.R. Heath, "Coglueing homotopy equivalences", Math.Z. 113 (1970) 313-325,
available here, which gives conditions for a pullback of homotopy equivalences to be a homotopy equivalence. A fibre is the pullback over a point. One uses the fact that any map, which in this instance is a fibration, factors through a homotopy equivalence and a fibration, as given by mland. The theorem is the dual of the gluing theorem for homotopy equivalences, which can be found in Topology and Groupoids, T&G, and has a dual proof. This proof gives also good control over the homotopies involved: see 7.4.2 (Addendum). The operation used is called "transport" in tom Dieck's book "Algebraic Topology", p.107 (EMS, 2008). It generalises the operation of the fundamental groupoid on higher homotopy groups.
A use of this "transport" in the fibration situation is also suggested in my sketch answer to this mathoverflow question.
answered Nov 18 '12 at 15:45
Ronnie BrownRonnie Brown
12.1k12939
$endgroup$
add a comment |
$begingroup$
The homotopy fiber of a map is only defined up to homotopy equivalence. I think this is a crucial point to your question.
One model for it is given by the following construction.
Let $f: X to Y$ be any map. Then we may factor it as a homotopy equivalence followed by a fibration, ie $X to Z to Y$, call the last map $g: Z to Y$ which is a fibration. Now one model for the homotopy fiber of $f$ is just the honest fiber of the map $g$.
In the case that $f$ itself is a fibration one may choose the obvious factorization to see that its fiber is one model for a homotopy fiber.
But since you asked for this very explicit model, at least assuming that all spaces are CW complexes, here is a brief idea of an argument. Comparing the long exact sequences of the fibration $f$ and the long exact sequence associated to the homotopy fiber of $f$ together with the "good choice" of a map from the fiber to the homotopy fiber and a 5 Lemma argument shows that fiber and homotopy fiber are weakly equivalent. By Whitehead then they are also homotopy equivalent.
By the good choice I mean the following:
A map from the fiber $F$ to $Ytimes_{X} X^{I}$ is given by compatible maps
$F to Y$ which you choose to be the inclusion and $F to X^{I}$ which can be chosen as the constant map at the basepoint of $X$.
answered Apr 14 '12 at 21:27
mlandmland
1,9791811
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
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active
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votes
$begingroup$
This is a consequence of the main theorem in
R. Brown and P.R. Heath, "Coglueing homotopy equivalences", Math.Z. 113 (1970) 313-325,
available here, which gives conditions for a pullback of homotopy equivalences to be a homotopy equivalence. A fibre is the pullback over a point. One uses the fact that any map, which in this instance is a fibration, factors through a homotopy equivalence and a fibration, as given by mland. The theorem is the dual of the gluing theorem for homotopy equivalences, which can be found in Topology and Groupoids, T&G, and has a dual proof. This proof gives also good control over the homotopies involved: see 7.4.2 (Addendum). The operation used is called "transport" in tom Dieck's book "Algebraic Topology", p.107 (EMS, 2008). It generalises the operation of the fundamental groupoid on higher homotopy groups.
A use of this "transport" in the fibration situation is also suggested in my sketch answer to this mathoverflow question.
answered Nov 18 '12 at 15:45
Ronnie BrownRonnie Brown
12.1k12939
$endgroup$
add a comment |
$begingroup$
This is a consequence of the main theorem in
R. Brown and P.R. Heath, "Coglueing homotopy equivalences", Math.Z. 113 (1970) 313-325,
available here, which gives conditions for a pullback of homotopy equivalences to be a homotopy equivalence. A fibre is the pullback over a point. One uses the fact that any map, which in this instance is a fibration, factors through a homotopy equivalence and a fibration, as given by mland. The theorem is the dual of the gluing theorem for homotopy equivalences, which can be found in Topology and Groupoids, T&G, and has a dual proof. This proof gives also good control over the homotopies involved: see 7.4.2 (Addendum). The operation used is called "transport" in tom Dieck's book "Algebraic Topology", p.107 (EMS, 2008). It generalises the operation of the fundamental groupoid on higher homotopy groups.
A use of this "transport" in the fibration situation is also suggested in my sketch answer to this mathoverflow question.
answered Nov 18 '12 at 15:45
Ronnie BrownRonnie Brown
12.1k12939
$endgroup$
add a comment |
$begingroup$
This is a consequence of the main theorem in
R. Brown and P.R. Heath, "Coglueing homotopy equivalences", Math.Z. 113 (1970) 313-325,
available here, which gives conditions for a pullback of homotopy equivalences to be a homotopy equivalence. A fibre is the pullback over a point. One uses the fact that any map, which in this instance is a fibration, factors through a homotopy equivalence and a fibration, as given by mland. The theorem is the dual of the gluing theorem for homotopy equivalences, which can be found in Topology and Groupoids, T&G, and has a dual proof. This proof gives also good control over the homotopies involved: see 7.4.2 (Addendum). The operation used is called "transport" in tom Dieck's book "Algebraic Topology", p.107 (EMS, 2008). It generalises the operation of the fundamental groupoid on higher homotopy groups.
A use of this "transport" in the fibration situation is also suggested in my sketch answer to this mathoverflow question.
answered Nov 18 '12 at 15:45
Ronnie BrownRonnie Brown
12.1k12939
$endgroup$
This is a consequence of the main theorem in
R. Brown and P.R. Heath, "Coglueing homotopy equivalences", Math.Z. 113 (1970) 313-325,
available here, which gives conditions for a pullback of homotopy equivalences to be a homotopy equivalence. A fibre is the pullback over a point. One uses the fact that any map, which in this instance is a fibration, factors through a homotopy equivalence and a fibration, as given by mland. The theorem is the dual of the gluing theorem for homotopy equivalences, which can be found in Topology and Groupoids, T&G, and has a dual proof. This proof gives also good control over the homotopies involved: see 7.4.2 (Addendum). The operation used is called "transport" in tom Dieck's book "Algebraic Topology", p.107 (EMS, 2008). It generalises the operation of the fundamental groupoid on higher homotopy groups.
A use of this "transport" in the fibration situation is also suggested in my sketch answer to this mathoverflow question.
answered Nov 18 '12 at 15:45
Ronnie BrownRonnie Brown
12.1k12939
edited Dec 25 '18 at 12:33
answered Nov 18 '12 at 15:45
Ronnie BrownRonnie Brown
12.1k12939
answered Nov 18 '12 at 15:45
Ronnie BrownRonnie Brown
12.1k12939
answered Nov 18 '12 at 15:45
Ronnie BrownRonnie Brown
12.1k12939
12.1k12939
add a comment |
add a comment |
$begingroup$
The homotopy fiber of a map is only defined up to homotopy equivalence. I think this is a crucial point to your question.
One model for it is given by the following construction.
Let $f: X to Y$ be any map. Then we may factor it as a homotopy equivalence followed by a fibration, ie $X to Z to Y$, call the last map $g: Z to Y$ which is a fibration. Now one model for the homotopy fiber of $f$ is just the honest fiber of the map $g$.
In the case that $f$ itself is a fibration one may choose the obvious factorization to see that its fiber is one model for a homotopy fiber.
But since you asked for this very explicit model, at least assuming that all spaces are CW complexes, here is a brief idea of an argument. Comparing the long exact sequences of the fibration $f$ and the long exact sequence associated to the homotopy fiber of $f$ together with the "good choice" of a map from the fiber to the homotopy fiber and a 5 Lemma argument shows that fiber and homotopy fiber are weakly equivalent. By Whitehead then they are also homotopy equivalent.
By the good choice I mean the following:
A map from the fiber $F$ to $Ytimes_{X} X^{I}$ is given by compatible maps
$F to Y$ which you choose to be the inclusion and $F to X^{I}$ which can be chosen as the constant map at the basepoint of $X$.
answered Apr 14 '12 at 21:27
mlandmland
1,9791811
$endgroup$
add a comment |
$begingroup$
The homotopy fiber of a map is only defined up to homotopy equivalence. I think this is a crucial point to your question.
One model for it is given by the following construction.
Let $f: X to Y$ be any map. Then we may factor it as a homotopy equivalence followed by a fibration, ie $X to Z to Y$, call the last map $g: Z to Y$ which is a fibration. Now one model for the homotopy fiber of $f$ is just the honest fiber of the map $g$.
In the case that $f$ itself is a fibration one may choose the obvious factorization to see that its fiber is one model for a homotopy fiber.
But since you asked for this very explicit model, at least assuming that all spaces are CW complexes, here is a brief idea of an argument. Comparing the long exact sequences of the fibration $f$ and the long exact sequence associated to the homotopy fiber of $f$ together with the "good choice" of a map from the fiber to the homotopy fiber and a 5 Lemma argument shows that fiber and homotopy fiber are weakly equivalent. By Whitehead then they are also homotopy equivalent.
By the good choice I mean the following:
A map from the fiber $F$ to $Ytimes_{X} X^{I}$ is given by compatible maps
$F to Y$ which you choose to be the inclusion and $F to X^{I}$ which can be chosen as the constant map at the basepoint of $X$.
answered Apr 14 '12 at 21:27
mlandmland
1,9791811
$endgroup$
add a comment |
$begingroup$
The homotopy fiber of a map is only defined up to homotopy equivalence. I think this is a crucial point to your question.
One model for it is given by the following construction.
Let $f: X to Y$ be any map. Then we may factor it as a homotopy equivalence followed by a fibration, ie $X to Z to Y$, call the last map $g: Z to Y$ which is a fibration. Now one model for the homotopy fiber of $f$ is just the honest fiber of the map $g$.
In the case that $f$ itself is a fibration one may choose the obvious factorization to see that its fiber is one model for a homotopy fiber.
But since you asked for this very explicit model, at least assuming that all spaces are CW complexes, here is a brief idea of an argument. Comparing the long exact sequences of the fibration $f$ and the long exact sequence associated to the homotopy fiber of $f$ together with the "good choice" of a map from the fiber to the homotopy fiber and a 5 Lemma argument shows that fiber and homotopy fiber are weakly equivalent. By Whitehead then they are also homotopy equivalent.
By the good choice I mean the following:
A map from the fiber $F$ to $Ytimes_{X} X^{I}$ is given by compatible maps
$F to Y$ which you choose to be the inclusion and $F to X^{I}$ which can be chosen as the constant map at the basepoint of $X$.
answered Apr 14 '12 at 21:27
mlandmland
1,9791811
$endgroup$
The homotopy fiber of a map is only defined up to homotopy equivalence. I think this is a crucial point to your question.
One model for it is given by the following construction.
Let $f: X to Y$ be any map. Then we may factor it as a homotopy equivalence followed by a fibration, ie $X to Z to Y$, call the last map $g: Z to Y$ which is a fibration. Now one model for the homotopy fiber of $f$ is just the honest fiber of the map $g$.
In the case that $f$ itself is a fibration one may choose the obvious factorization to see that its fiber is one model for a homotopy fiber.
But since you asked for this very explicit model, at least assuming that all spaces are CW complexes, here is a brief idea of an argument. Comparing the long exact sequences of the fibration $f$ and the long exact sequence associated to the homotopy fiber of $f$ together with the "good choice" of a map from the fiber to the homotopy fiber and a 5 Lemma argument shows that fiber and homotopy fiber are weakly equivalent. By Whitehead then they are also homotopy equivalent.
By the good choice I mean the following:
A map from the fiber $F$ to $Ytimes_{X} X^{I}$ is given by compatible maps
$F to Y$ which you choose to be the inclusion and $F to X^{I}$ which can be chosen as the constant map at the basepoint of $X$.
answered Apr 14 '12 at 21:27
mlandmland
1,9791811
answered Apr 14 '12 at 21:27
mlandmland
1,9791811
answered Apr 14 '12 at 21:27
mlandmland
1,9791811
answered Apr 14 '12 at 21:27
mlandmland
1,9791811
1,9791811
add a comment |
add a comment |
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1
$begingroup$
Not a hint, but a couple of arguments and references can be found in this MO-thread.
$endgroup$
– t.b.
May 10 '11 at 11:55
$begingroup$
Thanks, I already found this. But I don't want the complete argument.
$endgroup$
– user5262
May 10 '11 at 12:33
$begingroup$
The first step would be to construct a continuous function from one to the other. There appears to be a good choice from the fibre to the homotopy fibre.
$endgroup$
– Ryan Budney
May 10 '11 at 17:38