Homology of the fiber
$begingroup$
Let $f:Xrightarrow Y $ be a fibration (with fiber $F$) between simply connected spaces such that
$H_{ast}(f):H_{ast}(X,mathbb{Z})rightarrow H_{ast}(Y,mathbb{Z})$ is an isomorphism for $astleq n$
Is it true that the reduced homology of the fiber is $tilde{H}_{ast}(F,mathbb{Z})=0$ for $astleq n$?
at.algebraic-topology homotopy-theory
edited Mar 18 at 17:08
Peter Mortensen
1675
asked Mar 18 at 11:38
ParisParis
1154
$endgroup$
add a comment |
$begingroup$
Let $f:Xrightarrow Y $ be a fibration (with fiber $F$) between simply connected spaces such that
$H_{ast}(f):H_{ast}(X,mathbb{Z})rightarrow H_{ast}(Y,mathbb{Z})$ is an isomorphism for $astleq n$
Is it true that the reduced homology of the fiber is $tilde{H}_{ast}(F,mathbb{Z})=0$ for $astleq n$?
at.algebraic-topology homotopy-theory
edited Mar 18 at 17:08
Peter Mortensen
1675
asked Mar 18 at 11:38
ParisParis
1154
$endgroup$
6
$begingroup$
What about the Hopf fibration $f:mathbb{S}^3rightarrow mathbb{S}^2$ with fiber $mathbb{S}^1$? $H_1(f)$ is an isomorphism but $H_1(mathbb{S}^1)=mathbb{Z}$.
$endgroup$
– abx
Mar 18 at 12:04
$begingroup$
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
Mar 18 at 22:02
add a comment |
$begingroup$
Let $f:Xrightarrow Y $ be a fibration (with fiber $F$) between simply connected spaces such that
$H_{ast}(f):H_{ast}(X,mathbb{Z})rightarrow H_{ast}(Y,mathbb{Z})$ is an isomorphism for $astleq n$
Is it true that the reduced homology of the fiber is $tilde{H}_{ast}(F,mathbb{Z})=0$ for $astleq n$?
at.algebraic-topology homotopy-theory
edited Mar 18 at 17:08
Peter Mortensen
1675
asked Mar 18 at 11:38
ParisParis
1154
$endgroup$
Let $f:Xrightarrow Y $ be a fibration (with fiber $F$) between simply connected spaces such that
$H_{ast}(f):H_{ast}(X,mathbb{Z})rightarrow H_{ast}(Y,mathbb{Z})$ is an isomorphism for $astleq n$
Is it true that the reduced homology of the fiber is $tilde{H}_{ast}(F,mathbb{Z})=0$ for $astleq n$?
at.algebraic-topology homotopy-theory
at.algebraic-topology homotopy-theory
edited Mar 18 at 17:08
Peter Mortensen
1675
asked Mar 18 at 11:38
ParisParis
1154
edited Mar 18 at 17:08
Peter Mortensen
1675
asked Mar 18 at 11:38
ParisParis
1154
edited Mar 18 at 17:08
Peter Mortensen
1675
edited Mar 18 at 17:08
Peter Mortensen
1675
edited Mar 18 at 17:08
Peter Mortensen
1675
1675
asked Mar 18 at 11:38
ParisParis
1154
asked Mar 18 at 11:38
ParisParis
1154
asked Mar 18 at 11:38
ParisParis
1154
1154
6
$begingroup$
What about the Hopf fibration $f:mathbb{S}^3rightarrow mathbb{S}^2$ with fiber $mathbb{S}^1$? $H_1(f)$ is an isomorphism but $H_1(mathbb{S}^1)=mathbb{Z}$.
$endgroup$
– abx
Mar 18 at 12:04
$begingroup$
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
Mar 18 at 22:02
add a comment |
6
$begingroup$
What about the Hopf fibration $f:mathbb{S}^3rightarrow mathbb{S}^2$ with fiber $mathbb{S}^1$? $H_1(f)$ is an isomorphism but $H_1(mathbb{S}^1)=mathbb{Z}$.
$endgroup$
– abx
Mar 18 at 12:04
$begingroup$
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
Mar 18 at 22:02
6
6
$begingroup$
What about the Hopf fibration $f:mathbb{S}^3rightarrow mathbb{S}^2$ with fiber $mathbb{S}^1$? $H_1(f)$ is an isomorphism but $H_1(mathbb{S}^1)=mathbb{Z}$.
$endgroup$
– abx
Mar 18 at 12:04
$begingroup$
What about the Hopf fibration $f:mathbb{S}^3rightarrow mathbb{S}^2$ with fiber $mathbb{S}^1$? $H_1(f)$ is an isomorphism but $H_1(mathbb{S}^1)=mathbb{Z}$.
$endgroup$
– abx
Mar 18 at 12:04
$begingroup$
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
Mar 18 at 22:02
$begingroup$
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
Mar 18 at 22:02
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 Mar 18 at 19:24
ThiKu
6,33012137
answered Mar 18 at 12:14
Fernando MuroFernando Muro
11.9k23465
$endgroup$
add a comment |
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1 Answer
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$begingroup$
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 Mar 18 at 19:24
ThiKu
6,33012137
answered Mar 18 at 12:14
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 Mar 18 at 19:24
ThiKu
6,33012137
answered Mar 18 at 12:14
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 Mar 18 at 19:24
ThiKu
6,33012137
answered Mar 18 at 12:14
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 Mar 18 at 19:24
ThiKu
6,33012137
answered Mar 18 at 12:14
Fernando MuroFernando Muro
11.9k23465
edited Mar 18 at 19:24
ThiKu
6,33012137
edited Mar 18 at 19:24
ThiKu
6,33012137
edited Mar 18 at 19:24
ThiKu
6,33012137
6,33012137
answered Mar 18 at 12:14
Fernando MuroFernando Muro
11.9k23465
answered Mar 18 at 12:14
Fernando MuroFernando Muro
11.9k23465
answered Mar 18 at 12:14
Fernando MuroFernando Muro
11.9k23465
11.9k23465
add a comment |
add a comment |
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$begingroup$
What about the Hopf fibration $f:mathbb{S}^3rightarrow mathbb{S}^2$ with fiber $mathbb{S}^1$? $H_1(f)$ is an isomorphism but $H_1(mathbb{S}^1)=mathbb{Z}$.
$endgroup$
– abx
Mar 18 at 12:04
$begingroup$
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
Mar 18 at 22:02