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Necessary condition on homology group for a set to be contractible



The Next CEO of Stack OverflowAlgebraic TopologyWhat is the necessary and sufficient condition for a CW-complex to have its homology groups torsion-free?Fundamental group of the Poincaré Homology SphereNecessary condition for removing a simplex and changing homotopy type.Does trivial fundamental group imply contractible?Homology of Eilenberg-MacLane $K(pi,1)$ in terms of group homology and TorHomology of contractible spaceFundamental group generators of null homology manifoldsHomology group of $mathbbS^1 vee mathbbRP^2$ and covering spacesDo contractible homology manifolds have one end?










3












$begingroup$


We call a topological space is contractible iff it is homotopic to a point. Since homology group is homotopy invariant, we can see that under any abelian group as coefficients set, a topological space $(X, tau)$ has $H_1(X) = 0$ if $X$ is contractible.



Now, can we find a necessary condition on the homology group of $X$ that can imply X is contractible using some abelian groups as coefficients? The reason why I want to focus on $H_1(X)$ is because, if a space is not contractible, then there will be a 1-chain that can not be deformed to a point while a 2-face can always be deformed to a point.



I noticed that when using $mathbbQ$ as the coefficients, "$H_1(X) = 0$" can not imply $X$ is contractible. The conterexample is the projective plane of order 2, $mathbbP^2$. When using $mathbbZ$ as coefficients, then for any $n >= 2$, $S^n$ (the n-sphere) has homology 1-group equal to $0$ but all of them are not contractible.



Could anyone find an abelian group $G$ such that I can conclude "using $G$ as the coefficients set, $H_1(X) = 0$ implies $X$ is contractible"?
Furthermore, if no matter what coefficients set I use, $H_1(X)$ is always $0$, can I conclude that $X$ is contractible?










share|cite|improve this question









$endgroup$
















    3












    $begingroup$


    We call a topological space is contractible iff it is homotopic to a point. Since homology group is homotopy invariant, we can see that under any abelian group as coefficients set, a topological space $(X, tau)$ has $H_1(X) = 0$ if $X$ is contractible.



    Now, can we find a necessary condition on the homology group of $X$ that can imply X is contractible using some abelian groups as coefficients? The reason why I want to focus on $H_1(X)$ is because, if a space is not contractible, then there will be a 1-chain that can not be deformed to a point while a 2-face can always be deformed to a point.



    I noticed that when using $mathbbQ$ as the coefficients, "$H_1(X) = 0$" can not imply $X$ is contractible. The conterexample is the projective plane of order 2, $mathbbP^2$. When using $mathbbZ$ as coefficients, then for any $n >= 2$, $S^n$ (the n-sphere) has homology 1-group equal to $0$ but all of them are not contractible.



    Could anyone find an abelian group $G$ such that I can conclude "using $G$ as the coefficients set, $H_1(X) = 0$ implies $X$ is contractible"?
    Furthermore, if no matter what coefficients set I use, $H_1(X)$ is always $0$, can I conclude that $X$ is contractible?










    share|cite|improve this question









    $endgroup$














      3












      3








      3





      $begingroup$


      We call a topological space is contractible iff it is homotopic to a point. Since homology group is homotopy invariant, we can see that under any abelian group as coefficients set, a topological space $(X, tau)$ has $H_1(X) = 0$ if $X$ is contractible.



      Now, can we find a necessary condition on the homology group of $X$ that can imply X is contractible using some abelian groups as coefficients? The reason why I want to focus on $H_1(X)$ is because, if a space is not contractible, then there will be a 1-chain that can not be deformed to a point while a 2-face can always be deformed to a point.



      I noticed that when using $mathbbQ$ as the coefficients, "$H_1(X) = 0$" can not imply $X$ is contractible. The conterexample is the projective plane of order 2, $mathbbP^2$. When using $mathbbZ$ as coefficients, then for any $n >= 2$, $S^n$ (the n-sphere) has homology 1-group equal to $0$ but all of them are not contractible.



      Could anyone find an abelian group $G$ such that I can conclude "using $G$ as the coefficients set, $H_1(X) = 0$ implies $X$ is contractible"?
      Furthermore, if no matter what coefficients set I use, $H_1(X)$ is always $0$, can I conclude that $X$ is contractible?










      share|cite|improve this question









      $endgroup$




      We call a topological space is contractible iff it is homotopic to a point. Since homology group is homotopy invariant, we can see that under any abelian group as coefficients set, a topological space $(X, tau)$ has $H_1(X) = 0$ if $X$ is contractible.



      Now, can we find a necessary condition on the homology group of $X$ that can imply X is contractible using some abelian groups as coefficients? The reason why I want to focus on $H_1(X)$ is because, if a space is not contractible, then there will be a 1-chain that can not be deformed to a point while a 2-face can always be deformed to a point.



      I noticed that when using $mathbbQ$ as the coefficients, "$H_1(X) = 0$" can not imply $X$ is contractible. The conterexample is the projective plane of order 2, $mathbbP^2$. When using $mathbbZ$ as coefficients, then for any $n >= 2$, $S^n$ (the n-sphere) has homology 1-group equal to $0$ but all of them are not contractible.



      Could anyone find an abelian group $G$ such that I can conclude "using $G$ as the coefficients set, $H_1(X) = 0$ implies $X$ is contractible"?
      Furthermore, if no matter what coefficients set I use, $H_1(X)$ is always $0$, can I conclude that $X$ is contractible?







      algebraic-topology simplicial-complex






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      asked 1 hour ago









      Sanae KochiyaSanae Kochiya

      626




      626




















          3 Answers
          3






          active

          oldest

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          1












          $begingroup$

          The first homology group is far from enough to detect contractibility, since spaces can have non-vanishing higher homology groups.



          It's not even enough to have $H_n(X;G)$ vanish for every $n$ and $G$. For one thing there are spaces which are weakly contractible (i.e. all their homotopy vanish and hence their homology as well) but which are not contractible, like the Warsaw Circle.



          By Whitehead's Theorem a weakly contractible space which is not contractible cannot have the homotopy type of a CW complex, so we can ask if vanishing homology is enough to conclude that a CW complex is contractible. This still is not enough, because we can take the $2$-skeleton $S$ of the Poincare homology $3$-sphere, which is a finite $2$-dimensional CW complex whose homology groups vanish with any coefficients, but $pi_1(S)$ has order $120$ so it's not contractible.



          However there is an affirmative answer to your question that involves the fundamental group. If $X$ is a CW complex such that $pi_1(X) = 0$ and $H_n(X;mathbbZ)=0$ for $n > 1$, then it follows by Whitehead's Theorem and the Hurewicz Theorem that $X$ is contractible.






          share|cite|improve this answer









          $endgroup$




















            2












            $begingroup$

            A counterexample is the sphere $S^2$, whose first homology group will vanish for any coefficients, but which is not contractible (because its second homology group doesn't vanish).






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              Thank you for your response. Do you mind direct me to the proof of your statement?
              $endgroup$
              – Sanae Kochiya
              59 mins ago










            • $begingroup$
              For any abelian group of coefficients $A$, we have $H_1(S^2, A) = H_1(S^2, mathbbZ) otimes A$, e.g. by the universal coefficient theorem (since there's no torsion in the other homology groups).
              $endgroup$
              – hunter
              14 mins ago


















            0












            $begingroup$

            This is a very good question because this is exactly what early algebraic topologists cared about! The general case is no; there are no conditions on homology that are sufficient to say a space is contractible. The double comb space (https://topospaces.subwiki.org/wiki/Double_comb_space) is a space whose homology (and homotopy) groups are all trivial with coefficients in any group. It also is not contractible meaning it is not homotopy equivalent to a point.



            But when you have a great question, a counterexample should not dissuade you. Can we put restrictions on a space so that trivial homology (with coefficients in integers) implies it is contractible? The answer is yes. If we restrict to CW complexes, you can prove that any map that induces an isomorphism on all homotopy groups must be a homotopy equivalence. This is called Whitehead's theorem. One of its corollaries is that between simply connected CW complexes, any map that induces isomorphisms on homology groups is a homotopy equivalence. This means that a simply connected CW complex with trivial homology is contractible since the map to a point induces isomorphisms on homology.






            share|cite|improve this answer









            $endgroup$













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              3 Answers
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              3 Answers
              3






              active

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              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              The first homology group is far from enough to detect contractibility, since spaces can have non-vanishing higher homology groups.



              It's not even enough to have $H_n(X;G)$ vanish for every $n$ and $G$. For one thing there are spaces which are weakly contractible (i.e. all their homotopy vanish and hence their homology as well) but which are not contractible, like the Warsaw Circle.



              By Whitehead's Theorem a weakly contractible space which is not contractible cannot have the homotopy type of a CW complex, so we can ask if vanishing homology is enough to conclude that a CW complex is contractible. This still is not enough, because we can take the $2$-skeleton $S$ of the Poincare homology $3$-sphere, which is a finite $2$-dimensional CW complex whose homology groups vanish with any coefficients, but $pi_1(S)$ has order $120$ so it's not contractible.



              However there is an affirmative answer to your question that involves the fundamental group. If $X$ is a CW complex such that $pi_1(X) = 0$ and $H_n(X;mathbbZ)=0$ for $n > 1$, then it follows by Whitehead's Theorem and the Hurewicz Theorem that $X$ is contractible.






              share|cite|improve this answer









              $endgroup$

















                1












                $begingroup$

                The first homology group is far from enough to detect contractibility, since spaces can have non-vanishing higher homology groups.



                It's not even enough to have $H_n(X;G)$ vanish for every $n$ and $G$. For one thing there are spaces which are weakly contractible (i.e. all their homotopy vanish and hence their homology as well) but which are not contractible, like the Warsaw Circle.



                By Whitehead's Theorem a weakly contractible space which is not contractible cannot have the homotopy type of a CW complex, so we can ask if vanishing homology is enough to conclude that a CW complex is contractible. This still is not enough, because we can take the $2$-skeleton $S$ of the Poincare homology $3$-sphere, which is a finite $2$-dimensional CW complex whose homology groups vanish with any coefficients, but $pi_1(S)$ has order $120$ so it's not contractible.



                However there is an affirmative answer to your question that involves the fundamental group. If $X$ is a CW complex such that $pi_1(X) = 0$ and $H_n(X;mathbbZ)=0$ for $n > 1$, then it follows by Whitehead's Theorem and the Hurewicz Theorem that $X$ is contractible.






                share|cite|improve this answer









                $endgroup$















                  1












                  1








                  1





                  $begingroup$

                  The first homology group is far from enough to detect contractibility, since spaces can have non-vanishing higher homology groups.



                  It's not even enough to have $H_n(X;G)$ vanish for every $n$ and $G$. For one thing there are spaces which are weakly contractible (i.e. all their homotopy vanish and hence their homology as well) but which are not contractible, like the Warsaw Circle.



                  By Whitehead's Theorem a weakly contractible space which is not contractible cannot have the homotopy type of a CW complex, so we can ask if vanishing homology is enough to conclude that a CW complex is contractible. This still is not enough, because we can take the $2$-skeleton $S$ of the Poincare homology $3$-sphere, which is a finite $2$-dimensional CW complex whose homology groups vanish with any coefficients, but $pi_1(S)$ has order $120$ so it's not contractible.



                  However there is an affirmative answer to your question that involves the fundamental group. If $X$ is a CW complex such that $pi_1(X) = 0$ and $H_n(X;mathbbZ)=0$ for $n > 1$, then it follows by Whitehead's Theorem and the Hurewicz Theorem that $X$ is contractible.






                  share|cite|improve this answer









                  $endgroup$



                  The first homology group is far from enough to detect contractibility, since spaces can have non-vanishing higher homology groups.



                  It's not even enough to have $H_n(X;G)$ vanish for every $n$ and $G$. For one thing there are spaces which are weakly contractible (i.e. all their homotopy vanish and hence their homology as well) but which are not contractible, like the Warsaw Circle.



                  By Whitehead's Theorem a weakly contractible space which is not contractible cannot have the homotopy type of a CW complex, so we can ask if vanishing homology is enough to conclude that a CW complex is contractible. This still is not enough, because we can take the $2$-skeleton $S$ of the Poincare homology $3$-sphere, which is a finite $2$-dimensional CW complex whose homology groups vanish with any coefficients, but $pi_1(S)$ has order $120$ so it's not contractible.



                  However there is an affirmative answer to your question that involves the fundamental group. If $X$ is a CW complex such that $pi_1(X) = 0$ and $H_n(X;mathbbZ)=0$ for $n > 1$, then it follows by Whitehead's Theorem and the Hurewicz Theorem that $X$ is contractible.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 41 mins ago









                  WilliamWilliam

                  2,9351224




                  2,9351224





















                      2












                      $begingroup$

                      A counterexample is the sphere $S^2$, whose first homology group will vanish for any coefficients, but which is not contractible (because its second homology group doesn't vanish).






                      share|cite|improve this answer









                      $endgroup$












                      • $begingroup$
                        Thank you for your response. Do you mind direct me to the proof of your statement?
                        $endgroup$
                        – Sanae Kochiya
                        59 mins ago










                      • $begingroup$
                        For any abelian group of coefficients $A$, we have $H_1(S^2, A) = H_1(S^2, mathbbZ) otimes A$, e.g. by the universal coefficient theorem (since there's no torsion in the other homology groups).
                        $endgroup$
                        – hunter
                        14 mins ago















                      2












                      $begingroup$

                      A counterexample is the sphere $S^2$, whose first homology group will vanish for any coefficients, but which is not contractible (because its second homology group doesn't vanish).






                      share|cite|improve this answer









                      $endgroup$












                      • $begingroup$
                        Thank you for your response. Do you mind direct me to the proof of your statement?
                        $endgroup$
                        – Sanae Kochiya
                        59 mins ago










                      • $begingroup$
                        For any abelian group of coefficients $A$, we have $H_1(S^2, A) = H_1(S^2, mathbbZ) otimes A$, e.g. by the universal coefficient theorem (since there's no torsion in the other homology groups).
                        $endgroup$
                        – hunter
                        14 mins ago













                      2












                      2








                      2





                      $begingroup$

                      A counterexample is the sphere $S^2$, whose first homology group will vanish for any coefficients, but which is not contractible (because its second homology group doesn't vanish).






                      share|cite|improve this answer









                      $endgroup$



                      A counterexample is the sphere $S^2$, whose first homology group will vanish for any coefficients, but which is not contractible (because its second homology group doesn't vanish).







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 1 hour ago









                      hunterhunter

                      15.4k32640




                      15.4k32640











                      • $begingroup$
                        Thank you for your response. Do you mind direct me to the proof of your statement?
                        $endgroup$
                        – Sanae Kochiya
                        59 mins ago










                      • $begingroup$
                        For any abelian group of coefficients $A$, we have $H_1(S^2, A) = H_1(S^2, mathbbZ) otimes A$, e.g. by the universal coefficient theorem (since there's no torsion in the other homology groups).
                        $endgroup$
                        – hunter
                        14 mins ago
















                      • $begingroup$
                        Thank you for your response. Do you mind direct me to the proof of your statement?
                        $endgroup$
                        – Sanae Kochiya
                        59 mins ago










                      • $begingroup$
                        For any abelian group of coefficients $A$, we have $H_1(S^2, A) = H_1(S^2, mathbbZ) otimes A$, e.g. by the universal coefficient theorem (since there's no torsion in the other homology groups).
                        $endgroup$
                        – hunter
                        14 mins ago















                      $begingroup$
                      Thank you for your response. Do you mind direct me to the proof of your statement?
                      $endgroup$
                      – Sanae Kochiya
                      59 mins ago




                      $begingroup$
                      Thank you for your response. Do you mind direct me to the proof of your statement?
                      $endgroup$
                      – Sanae Kochiya
                      59 mins ago












                      $begingroup$
                      For any abelian group of coefficients $A$, we have $H_1(S^2, A) = H_1(S^2, mathbbZ) otimes A$, e.g. by the universal coefficient theorem (since there's no torsion in the other homology groups).
                      $endgroup$
                      – hunter
                      14 mins ago




                      $begingroup$
                      For any abelian group of coefficients $A$, we have $H_1(S^2, A) = H_1(S^2, mathbbZ) otimes A$, e.g. by the universal coefficient theorem (since there's no torsion in the other homology groups).
                      $endgroup$
                      – hunter
                      14 mins ago











                      0












                      $begingroup$

                      This is a very good question because this is exactly what early algebraic topologists cared about! The general case is no; there are no conditions on homology that are sufficient to say a space is contractible. The double comb space (https://topospaces.subwiki.org/wiki/Double_comb_space) is a space whose homology (and homotopy) groups are all trivial with coefficients in any group. It also is not contractible meaning it is not homotopy equivalent to a point.



                      But when you have a great question, a counterexample should not dissuade you. Can we put restrictions on a space so that trivial homology (with coefficients in integers) implies it is contractible? The answer is yes. If we restrict to CW complexes, you can prove that any map that induces an isomorphism on all homotopy groups must be a homotopy equivalence. This is called Whitehead's theorem. One of its corollaries is that between simply connected CW complexes, any map that induces isomorphisms on homology groups is a homotopy equivalence. This means that a simply connected CW complex with trivial homology is contractible since the map to a point induces isomorphisms on homology.






                      share|cite|improve this answer









                      $endgroup$

















                        0












                        $begingroup$

                        This is a very good question because this is exactly what early algebraic topologists cared about! The general case is no; there are no conditions on homology that are sufficient to say a space is contractible. The double comb space (https://topospaces.subwiki.org/wiki/Double_comb_space) is a space whose homology (and homotopy) groups are all trivial with coefficients in any group. It also is not contractible meaning it is not homotopy equivalent to a point.



                        But when you have a great question, a counterexample should not dissuade you. Can we put restrictions on a space so that trivial homology (with coefficients in integers) implies it is contractible? The answer is yes. If we restrict to CW complexes, you can prove that any map that induces an isomorphism on all homotopy groups must be a homotopy equivalence. This is called Whitehead's theorem. One of its corollaries is that between simply connected CW complexes, any map that induces isomorphisms on homology groups is a homotopy equivalence. This means that a simply connected CW complex with trivial homology is contractible since the map to a point induces isomorphisms on homology.






                        share|cite|improve this answer









                        $endgroup$















                          0












                          0








                          0





                          $begingroup$

                          This is a very good question because this is exactly what early algebraic topologists cared about! The general case is no; there are no conditions on homology that are sufficient to say a space is contractible. The double comb space (https://topospaces.subwiki.org/wiki/Double_comb_space) is a space whose homology (and homotopy) groups are all trivial with coefficients in any group. It also is not contractible meaning it is not homotopy equivalent to a point.



                          But when you have a great question, a counterexample should not dissuade you. Can we put restrictions on a space so that trivial homology (with coefficients in integers) implies it is contractible? The answer is yes. If we restrict to CW complexes, you can prove that any map that induces an isomorphism on all homotopy groups must be a homotopy equivalence. This is called Whitehead's theorem. One of its corollaries is that between simply connected CW complexes, any map that induces isomorphisms on homology groups is a homotopy equivalence. This means that a simply connected CW complex with trivial homology is contractible since the map to a point induces isomorphisms on homology.






                          share|cite|improve this answer









                          $endgroup$



                          This is a very good question because this is exactly what early algebraic topologists cared about! The general case is no; there are no conditions on homology that are sufficient to say a space is contractible. The double comb space (https://topospaces.subwiki.org/wiki/Double_comb_space) is a space whose homology (and homotopy) groups are all trivial with coefficients in any group. It also is not contractible meaning it is not homotopy equivalent to a point.



                          But when you have a great question, a counterexample should not dissuade you. Can we put restrictions on a space so that trivial homology (with coefficients in integers) implies it is contractible? The answer is yes. If we restrict to CW complexes, you can prove that any map that induces an isomorphism on all homotopy groups must be a homotopy equivalence. This is called Whitehead's theorem. One of its corollaries is that between simply connected CW complexes, any map that induces isomorphisms on homology groups is a homotopy equivalence. This means that a simply connected CW complex with trivial homology is contractible since the map to a point induces isomorphisms on homology.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 35 mins ago









                          Connor MalinConnor Malin

                          584111




                          584111



























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