Homology of the fiber












9












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
    17 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
    7 hours ago
















9












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
    17 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
    7 hours ago














9












9








9


2



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 12 hours ago









Peter Mortensen

1675




1675










asked 17 hours ago









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
    17 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
    7 hours ago














  • 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
    17 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
    7 hours ago








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
17 hours ago




$begingroup$
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
17 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
7 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
7 hours ago










1 Answer
1






active

oldest

votes


















21












$begingroup$

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.






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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.






    share|cite|improve this answer











    $endgroup$


















      21












      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$
















        21












        21








        21





        $begingroup$

        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.






        share|cite|improve this answer











        $endgroup$



        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 9 hours ago









        ThiKu

        6,32012137




        6,32012137










        answered 17 hours ago









        Fernando MuroFernando Muro

        11.9k23465




        11.9k23465






























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