Homology of the fiberSpectral sequence for reduced homologyweak equivalence of simplicial setsHomology of loop spaceIs the suspension of a weak equivalence again a weak equivalence?localization and $E_infty$-spacesfiber, homotopy fiber of spacesVietoris-Begle theorem for simplicial setsPullback and homologyTo compare the total, base and fiber spaces of two fiber bundleshomology of a base space of a a fiber sequence
Homology of the fiber
Spectral sequence for reduced homologyweak equivalence of simplicial setsHomology of loop spaceIs the suspension of a weak equivalence again a weak equivalence?localization and $E_infty$-spacesfiber, homotopy fiber of spacesVietoris-Begle theorem for simplicial setsPullback and homologyTo compare the total, base and fiber spaces of two fiber bundleshomology of a base space of a a fiber sequence
$begingroup$
Let $f:Xrightarrow Y $ be a fibration (with fiber $F$) between simply connected spaces such that
$H_ast(f):H_ast(X,mathbbZ)rightarrow H_ast(Y,mathbbZ)$ is an isomorphism for $astleq n$
Is it true that the reduced homology of the fiber is $tildeH_ast(F,mathbbZ)=0$ for $astleq n$?
at.algebraic-topology homotopy-theory
edited 8 hours ago
Peter Mortensen
1675
asked 14 hours ago
ParisParis
1154
$endgroup$
add a comment |
$begingroup$
Let $f:Xrightarrow Y $ be a fibration (with fiber $F$) between simply connected spaces such that
$H_ast(f):H_ast(X,mathbbZ)rightarrow H_ast(Y,mathbbZ)$ is an isomorphism for $astleq n$
Is it true that the reduced homology of the fiber is $tildeH_ast(F,mathbbZ)=0$ for $astleq n$?
at.algebraic-topology homotopy-theory
edited 8 hours ago
Peter Mortensen
1675
asked 14 hours ago
ParisParis
1154
$endgroup$
6
$begingroup$
What about the Hopf fibration $f:mathbbS^3rightarrow mathbbS^2$ with fiber $mathbbS^1$? $H_1(f)$ is an isomorphism but $H_1(mathbbS^1)=mathbbZ$.
$endgroup$
– abx
13 hours ago
$begingroup$
Besides the proof below, this (the vanishing of the reduced homology of fiber below dimension n) also admits an easy proof using the Serre spectral sequence.
$endgroup$
– Nicholas Kuhn
3 hours ago
add a comment |
$begingroup$
Let $f:Xrightarrow Y $ be a fibration (with fiber $F$) between simply connected spaces such that
$H_ast(f):H_ast(X,mathbbZ)rightarrow H_ast(Y,mathbbZ)$ is an isomorphism for $astleq n$
Is it true that the reduced homology of the fiber is $tildeH_ast(F,mathbbZ)=0$ for $astleq n$?
at.algebraic-topology homotopy-theory
edited 8 hours ago
Peter Mortensen
1675
asked 14 hours ago
ParisParis
1154
$endgroup$
Let $f:Xrightarrow Y $ be a fibration (with fiber $F$) between simply connected spaces such that
$H_ast(f):H_ast(X,mathbbZ)rightarrow H_ast(Y,mathbbZ)$ is an isomorphism for $astleq n$
Is it true that the reduced homology of the fiber is $tildeH_ast(F,mathbbZ)=0$ for $astleq n$?
at.algebraic-topology homotopy-theory
at.algebraic-topology homotopy-theory
edited 8 hours ago
Peter Mortensen
1675
asked 14 hours ago
ParisParis
1154
edited 8 hours ago
Peter Mortensen
1675
asked 14 hours ago
ParisParis
1154
edited 8 hours ago
Peter Mortensen
1675
edited 8 hours ago
Peter Mortensen
1675
edited 8 hours ago
Peter Mortensen
1675
1675
asked 14 hours ago
ParisParis
1154
asked 14 hours ago
ParisParis
1154
asked 14 hours ago
ParisParis
1154
1154
6
$begingroup$
What about the Hopf fibration $f:mathbbS^3rightarrow mathbbS^2$ with fiber $mathbbS^1$? $H_1(f)$ is an isomorphism but $H_1(mathbbS^1)=mathbbZ$.
$endgroup$
– abx
13 hours ago
$begingroup$
Besides the proof below, this (the vanishing of the reduced homology of fiber below dimension n) also admits an easy proof using the Serre spectral sequence.
$endgroup$
– Nicholas Kuhn
3 hours ago
add a comment |
6
$begingroup$
What about the Hopf fibration $f:mathbbS^3rightarrow mathbbS^2$ with fiber $mathbbS^1$? $H_1(f)$ is an isomorphism but $H_1(mathbbS^1)=mathbbZ$.
$endgroup$
– abx
13 hours ago
$begingroup$
Besides the proof below, this (the vanishing of the reduced homology of fiber below dimension n) also admits an easy proof using the Serre spectral sequence.
$endgroup$
– Nicholas Kuhn
3 hours ago
6
6
$begingroup$
What about the Hopf fibration $f:mathbbS^3rightarrow mathbbS^2$ with fiber $mathbbS^1$? $H_1(f)$ is an isomorphism but $H_1(mathbbS^1)=mathbbZ$.
$endgroup$
– abx
13 hours ago
$begingroup$
What about the Hopf fibration $f:mathbbS^3rightarrow mathbbS^2$ with fiber $mathbbS^1$? $H_1(f)$ is an isomorphism but $H_1(mathbbS^1)=mathbbZ$.
$endgroup$
– abx
13 hours ago
$begingroup$
Besides the proof below, this (the vanishing of the reduced homology of fiber below dimension n) also admits an easy proof using the Serre spectral sequence.
$endgroup$
– Nicholas Kuhn
3 hours ago
$begingroup$
Besides the proof below, this (the vanishing of the reduced homology of fiber below dimension n) also admits an easy proof using the Serre spectral sequence.
$endgroup$
– Nicholas Kuhn
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As usual, there's no loss of generality in assuming that $f$ is the inclusion of a subspace $Xsubset Y$, replacing $Y$ with the homotopy equivalent mapping cylinder of $f$ if necessary. By your assumptions and the five lemma, $H_*(Y,X)=0$ for $*leq n$, and the pair $(Y,X)$ is simply connected, therefore by the Hurewicz theorems $pi_*(Y,X)=0$ for $*leq n$. If $F$ denotes the homotopy fiber of $f$, then $pi_*(Y,X)=pi_*-1(F)$ in all dimensions, hence the previous computation ensures that $F$ is $(n-1)$-connected, so $H_*(F)=0$ for $*leq n-1$. As @abx shows in the comment above, in general $H_n(F)$ won't be trivial. The higher-dimensional Hopf fibrations provide further counterexamples, where even the fiber is simply connected.
edited 6 hours ago
ThiKu
6,32012137
answered 13 hours ago
Fernando MuroFernando Muro
11.9k23465
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As usual, there's no loss of generality in assuming that $f$ is the inclusion of a subspace $Xsubset Y$, replacing $Y$ with the homotopy equivalent mapping cylinder of $f$ if necessary. By your assumptions and the five lemma, $H_*(Y,X)=0$ for $*leq n$, and the pair $(Y,X)$ is simply connected, therefore by the Hurewicz theorems $pi_*(Y,X)=0$ for $*leq n$. If $F$ denotes the homotopy fiber of $f$, then $pi_*(Y,X)=pi_*-1(F)$ in all dimensions, hence the previous computation ensures that $F$ is $(n-1)$-connected, so $H_*(F)=0$ for $*leq n-1$. As @abx shows in the comment above, in general $H_n(F)$ won't be trivial. The higher-dimensional Hopf fibrations provide further counterexamples, where even the fiber is simply connected.
edited 6 hours ago
ThiKu
6,32012137
answered 13 hours ago
Fernando MuroFernando Muro
11.9k23465
$endgroup$
add a comment |
$begingroup$
As usual, there's no loss of generality in assuming that $f$ is the inclusion of a subspace $Xsubset Y$, replacing $Y$ with the homotopy equivalent mapping cylinder of $f$ if necessary. By your assumptions and the five lemma, $H_*(Y,X)=0$ for $*leq n$, and the pair $(Y,X)$ is simply connected, therefore by the Hurewicz theorems $pi_*(Y,X)=0$ for $*leq n$. If $F$ denotes the homotopy fiber of $f$, then $pi_*(Y,X)=pi_*-1(F)$ in all dimensions, hence the previous computation ensures that $F$ is $(n-1)$-connected, so $H_*(F)=0$ for $*leq n-1$. As @abx shows in the comment above, in general $H_n(F)$ won't be trivial. The higher-dimensional Hopf fibrations provide further counterexamples, where even the fiber is simply connected.
edited 6 hours ago
ThiKu
6,32012137
answered 13 hours ago
Fernando MuroFernando Muro
11.9k23465
$endgroup$
add a comment |
$begingroup$
As usual, there's no loss of generality in assuming that $f$ is the inclusion of a subspace $Xsubset Y$, replacing $Y$ with the homotopy equivalent mapping cylinder of $f$ if necessary. By your assumptions and the five lemma, $H_*(Y,X)=0$ for $*leq n$, and the pair $(Y,X)$ is simply connected, therefore by the Hurewicz theorems $pi_*(Y,X)=0$ for $*leq n$. If $F$ denotes the homotopy fiber of $f$, then $pi_*(Y,X)=pi_*-1(F)$ in all dimensions, hence the previous computation ensures that $F$ is $(n-1)$-connected, so $H_*(F)=0$ for $*leq n-1$. As @abx shows in the comment above, in general $H_n(F)$ won't be trivial. The higher-dimensional Hopf fibrations provide further counterexamples, where even the fiber is simply connected.
edited 6 hours ago
ThiKu
6,32012137
answered 13 hours ago
Fernando MuroFernando Muro
11.9k23465
$endgroup$
As usual, there's no loss of generality in assuming that $f$ is the inclusion of a subspace $Xsubset Y$, replacing $Y$ with the homotopy equivalent mapping cylinder of $f$ if necessary. By your assumptions and the five lemma, $H_*(Y,X)=0$ for $*leq n$, and the pair $(Y,X)$ is simply connected, therefore by the Hurewicz theorems $pi_*(Y,X)=0$ for $*leq n$. If $F$ denotes the homotopy fiber of $f$, then $pi_*(Y,X)=pi_*-1(F)$ in all dimensions, hence the previous computation ensures that $F$ is $(n-1)$-connected, so $H_*(F)=0$ for $*leq n-1$. As @abx shows in the comment above, in general $H_n(F)$ won't be trivial. The higher-dimensional Hopf fibrations provide further counterexamples, where even the fiber is simply connected.
edited 6 hours ago
ThiKu
6,32012137
answered 13 hours ago
Fernando MuroFernando Muro
11.9k23465
edited 6 hours ago
ThiKu
6,32012137
edited 6 hours ago
ThiKu
6,32012137
edited 6 hours ago
ThiKu
6,32012137
6,32012137
answered 13 hours ago
Fernando MuroFernando Muro
11.9k23465
answered 13 hours ago
Fernando MuroFernando Muro
11.9k23465
answered 13 hours ago
Fernando MuroFernando Muro
11.9k23465
11.9k23465
add a comment |
add a comment |
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$begingroup$
What about the Hopf fibration $f:mathbbS^3rightarrow mathbbS^2$ with fiber $mathbbS^1$? $H_1(f)$ is an isomorphism but $H_1(mathbbS^1)=mathbbZ$.
$endgroup$
– abx
13 hours ago
$begingroup$
Besides the proof below, this (the vanishing of the reduced homology of fiber below dimension n) also admits an easy proof using the Serre spectral sequence.
$endgroup$
– Nicholas Kuhn
3 hours ago