Construct compact space with some homology group not finitely generated












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I would like to construct a compact space $X$ that has a non finitely generated singular homology group $H_n (X)$ for some $n$. I thought about taking a countable wedge sum of 1-spheres, but this space is not compact. Another idea would be the Hawaiian earring which should be compact and have an infinitely generated homology group, but this is quite difficult to calculate.






algebraic-topology homology-cohomology







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    This thread might interest you: math.stackexchange.com/questions/2512580/…
    – Aleksandar Milivojevic
    Dec 2 '18 at 20:07
















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I would like to construct a compact space $X$ that has a non finitely generated singular homology group $H_n (X)$ for some $n$. I thought about taking a countable wedge sum of 1-spheres, but this space is not compact. Another idea would be the Hawaiian earring which should be compact and have an infinitely generated homology group, but this is quite difficult to calculate.






algebraic-topology homology-cohomology







share|cite|improve this question


















  • 1




    This thread might interest you: math.stackexchange.com/questions/2512580/…
    – Aleksandar Milivojevic
    Dec 2 '18 at 20:07














2












2








2







I would like to construct a compact space $X$ that has a non finitely generated singular homology group $H_n (X)$ for some $n$. I thought about taking a countable wedge sum of 1-spheres, but this space is not compact. Another idea would be the Hawaiian earring which should be compact and have an infinitely generated homology group, but this is quite difficult to calculate.






algebraic-topology homology-cohomology







share|cite|improve this question













I would like to construct a compact space $X$ that has a non finitely generated singular homology group $H_n (X)$ for some $n$. I thought about taking a countable wedge sum of 1-spheres, but this space is not compact. Another idea would be the Hawaiian earring which should be compact and have an infinitely generated homology group, but this is quite difficult to calculate.






algebraic-topology homology-cohomology




algebraic-topology homology-cohomology






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asked Dec 2 '18 at 19:41









CHwC

246111




246111








  • 1




    This thread might interest you: math.stackexchange.com/questions/2512580/…
    – Aleksandar Milivojevic
    Dec 2 '18 at 20:07














  • 1




    This thread might interest you: math.stackexchange.com/questions/2512580/…
    – Aleksandar Milivojevic
    Dec 2 '18 at 20:07








1




1




This thread might interest you: math.stackexchange.com/questions/2512580/…
– Aleksandar Milivojevic
Dec 2 '18 at 20:07




This thread might interest you: math.stackexchange.com/questions/2512580/…
– Aleksandar Milivojevic
Dec 2 '18 at 20:07










2 Answers
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The Hawaiian earring works just fine. You don't have to explicitly compute its homology to show it is not finitely generated. Specifically let $X$ be the Hawaiian earring space. Note that for any $ninmathbb{N}$, $X$ retracts onto a wedge of $n$ circles (just take $n$ of the circles that make up $X$, and map all the rest of the circles to the point where the circles meet). This implies that $H_1(X)$ has $mathbb{Z}^n$ as a direct summand for all $ninmathbb{Z}$. This implies $H_1(X)$ is not finitely generated.



For an even easier example, you could take $X$ to be any infinite compact totally disconnected space (say, ${0}cup{1/n:ninmathbb{Z}_+}$, or a Cantor set). Then $H_0(X)$ is not finitely generated, since it is freely generated by the path-components of $X$ and there are infinitely many path-components.






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  • how do you conclude that Z^n has to be a direct summand and the step where you say that it is freely generated by the path-components?
    – CHwC
    Dec 2 '18 at 20:48










  • okay the second point is simply the definition of singular homology, but what about the first?
    – CHwC
    Dec 2 '18 at 21:21










  • The retraction gives you a surjection of $H_1$ onto $mathbb{Z}^n$. Since $mathbb{Z}^n$ is free abelian, this surjects splits, and the claim follows by the splitting lemma en.m.wikipedia.org/wiki/Splitting_lemma
    – Aleksandar Milivojevic
    Dec 2 '18 at 22:01



















4














The Cantor set $C subset [0,1]$ has uncountably many path components (in fact, each single point subset is a path component of $C$). Hence $H_0(C)$ is a free abelian group with uncountably many generators.



Taking the suspension $Sigma$ and noting that $tilde{H}_{n+1}(Sigma X) approx tilde{H}_n(X)$, where $tilde{H}_*$ denotes reduced homology, you can construct examples for all $H_i$ with $i ge 0$. Recall that $tilde{H}_i = H_i$ for $i > 0$.



Edited: Instead of $C$ you can take any space $X$ with infinitely many path components. If you take $X = { 0 } cup { 1/n mid n in mathbb{N} }$ as in Eric Wofsey's answer, you get $Sigma X$ = Hawaiian earring.






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    2 Answers
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    4














    The Hawaiian earring works just fine. You don't have to explicitly compute its homology to show it is not finitely generated. Specifically let $X$ be the Hawaiian earring space. Note that for any $ninmathbb{N}$, $X$ retracts onto a wedge of $n$ circles (just take $n$ of the circles that make up $X$, and map all the rest of the circles to the point where the circles meet). This implies that $H_1(X)$ has $mathbb{Z}^n$ as a direct summand for all $ninmathbb{Z}$. This implies $H_1(X)$ is not finitely generated.



    For an even easier example, you could take $X$ to be any infinite compact totally disconnected space (say, ${0}cup{1/n:ninmathbb{Z}_+}$, or a Cantor set). Then $H_0(X)$ is not finitely generated, since it is freely generated by the path-components of $X$ and there are infinitely many path-components.






    share|cite|improve this answer





















    • how do you conclude that Z^n has to be a direct summand and the step where you say that it is freely generated by the path-components?
      – CHwC
      Dec 2 '18 at 20:48










    • okay the second point is simply the definition of singular homology, but what about the first?
      – CHwC
      Dec 2 '18 at 21:21










    • The retraction gives you a surjection of $H_1$ onto $mathbb{Z}^n$. Since $mathbb{Z}^n$ is free abelian, this surjects splits, and the claim follows by the splitting lemma en.m.wikipedia.org/wiki/Splitting_lemma
      – Aleksandar Milivojevic
      Dec 2 '18 at 22:01
















    4














    The Hawaiian earring works just fine. You don't have to explicitly compute its homology to show it is not finitely generated. Specifically let $X$ be the Hawaiian earring space. Note that for any $ninmathbb{N}$, $X$ retracts onto a wedge of $n$ circles (just take $n$ of the circles that make up $X$, and map all the rest of the circles to the point where the circles meet). This implies that $H_1(X)$ has $mathbb{Z}^n$ as a direct summand for all $ninmathbb{Z}$. This implies $H_1(X)$ is not finitely generated.



    For an even easier example, you could take $X$ to be any infinite compact totally disconnected space (say, ${0}cup{1/n:ninmathbb{Z}_+}$, or a Cantor set). Then $H_0(X)$ is not finitely generated, since it is freely generated by the path-components of $X$ and there are infinitely many path-components.






    share|cite|improve this answer





















    • how do you conclude that Z^n has to be a direct summand and the step where you say that it is freely generated by the path-components?
      – CHwC
      Dec 2 '18 at 20:48










    • okay the second point is simply the definition of singular homology, but what about the first?
      – CHwC
      Dec 2 '18 at 21:21










    • The retraction gives you a surjection of $H_1$ onto $mathbb{Z}^n$. Since $mathbb{Z}^n$ is free abelian, this surjects splits, and the claim follows by the splitting lemma en.m.wikipedia.org/wiki/Splitting_lemma
      – Aleksandar Milivojevic
      Dec 2 '18 at 22:01














    4












    4








    4






    The Hawaiian earring works just fine. You don't have to explicitly compute its homology to show it is not finitely generated. Specifically let $X$ be the Hawaiian earring space. Note that for any $ninmathbb{N}$, $X$ retracts onto a wedge of $n$ circles (just take $n$ of the circles that make up $X$, and map all the rest of the circles to the point where the circles meet). This implies that $H_1(X)$ has $mathbb{Z}^n$ as a direct summand for all $ninmathbb{Z}$. This implies $H_1(X)$ is not finitely generated.



    For an even easier example, you could take $X$ to be any infinite compact totally disconnected space (say, ${0}cup{1/n:ninmathbb{Z}_+}$, or a Cantor set). Then $H_0(X)$ is not finitely generated, since it is freely generated by the path-components of $X$ and there are infinitely many path-components.






    share|cite|improve this answer












    The Hawaiian earring works just fine. You don't have to explicitly compute its homology to show it is not finitely generated. Specifically let $X$ be the Hawaiian earring space. Note that for any $ninmathbb{N}$, $X$ retracts onto a wedge of $n$ circles (just take $n$ of the circles that make up $X$, and map all the rest of the circles to the point where the circles meet). This implies that $H_1(X)$ has $mathbb{Z}^n$ as a direct summand for all $ninmathbb{Z}$. This implies $H_1(X)$ is not finitely generated.



    For an even easier example, you could take $X$ to be any infinite compact totally disconnected space (say, ${0}cup{1/n:ninmathbb{Z}_+}$, or a Cantor set). Then $H_0(X)$ is not finitely generated, since it is freely generated by the path-components of $X$ and there are infinitely many path-components.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 2 '18 at 20:05









    Eric Wofsey

    180k12207335




    180k12207335












    • how do you conclude that Z^n has to be a direct summand and the step where you say that it is freely generated by the path-components?
      – CHwC
      Dec 2 '18 at 20:48










    • okay the second point is simply the definition of singular homology, but what about the first?
      – CHwC
      Dec 2 '18 at 21:21










    • The retraction gives you a surjection of $H_1$ onto $mathbb{Z}^n$. Since $mathbb{Z}^n$ is free abelian, this surjects splits, and the claim follows by the splitting lemma en.m.wikipedia.org/wiki/Splitting_lemma
      – Aleksandar Milivojevic
      Dec 2 '18 at 22:01


















    • how do you conclude that Z^n has to be a direct summand and the step where you say that it is freely generated by the path-components?
      – CHwC
      Dec 2 '18 at 20:48










    • okay the second point is simply the definition of singular homology, but what about the first?
      – CHwC
      Dec 2 '18 at 21:21










    • The retraction gives you a surjection of $H_1$ onto $mathbb{Z}^n$. Since $mathbb{Z}^n$ is free abelian, this surjects splits, and the claim follows by the splitting lemma en.m.wikipedia.org/wiki/Splitting_lemma
      – Aleksandar Milivojevic
      Dec 2 '18 at 22:01
















    how do you conclude that Z^n has to be a direct summand and the step where you say that it is freely generated by the path-components?
    – CHwC
    Dec 2 '18 at 20:48




    how do you conclude that Z^n has to be a direct summand and the step where you say that it is freely generated by the path-components?
    – CHwC
    Dec 2 '18 at 20:48












    okay the second point is simply the definition of singular homology, but what about the first?
    – CHwC
    Dec 2 '18 at 21:21




    okay the second point is simply the definition of singular homology, but what about the first?
    – CHwC
    Dec 2 '18 at 21:21












    The retraction gives you a surjection of $H_1$ onto $mathbb{Z}^n$. Since $mathbb{Z}^n$ is free abelian, this surjects splits, and the claim follows by the splitting lemma en.m.wikipedia.org/wiki/Splitting_lemma
    – Aleksandar Milivojevic
    Dec 2 '18 at 22:01




    The retraction gives you a surjection of $H_1$ onto $mathbb{Z}^n$. Since $mathbb{Z}^n$ is free abelian, this surjects splits, and the claim follows by the splitting lemma en.m.wikipedia.org/wiki/Splitting_lemma
    – Aleksandar Milivojevic
    Dec 2 '18 at 22:01











    4














    The Cantor set $C subset [0,1]$ has uncountably many path components (in fact, each single point subset is a path component of $C$). Hence $H_0(C)$ is a free abelian group with uncountably many generators.



    Taking the suspension $Sigma$ and noting that $tilde{H}_{n+1}(Sigma X) approx tilde{H}_n(X)$, where $tilde{H}_*$ denotes reduced homology, you can construct examples for all $H_i$ with $i ge 0$. Recall that $tilde{H}_i = H_i$ for $i > 0$.



    Edited: Instead of $C$ you can take any space $X$ with infinitely many path components. If you take $X = { 0 } cup { 1/n mid n in mathbb{N} }$ as in Eric Wofsey's answer, you get $Sigma X$ = Hawaiian earring.






    share|cite|improve this answer




























      4














      The Cantor set $C subset [0,1]$ has uncountably many path components (in fact, each single point subset is a path component of $C$). Hence $H_0(C)$ is a free abelian group with uncountably many generators.



      Taking the suspension $Sigma$ and noting that $tilde{H}_{n+1}(Sigma X) approx tilde{H}_n(X)$, where $tilde{H}_*$ denotes reduced homology, you can construct examples for all $H_i$ with $i ge 0$. Recall that $tilde{H}_i = H_i$ for $i > 0$.



      Edited: Instead of $C$ you can take any space $X$ with infinitely many path components. If you take $X = { 0 } cup { 1/n mid n in mathbb{N} }$ as in Eric Wofsey's answer, you get $Sigma X$ = Hawaiian earring.






      share|cite|improve this answer


























        4












        4








        4






        The Cantor set $C subset [0,1]$ has uncountably many path components (in fact, each single point subset is a path component of $C$). Hence $H_0(C)$ is a free abelian group with uncountably many generators.



        Taking the suspension $Sigma$ and noting that $tilde{H}_{n+1}(Sigma X) approx tilde{H}_n(X)$, where $tilde{H}_*$ denotes reduced homology, you can construct examples for all $H_i$ with $i ge 0$. Recall that $tilde{H}_i = H_i$ for $i > 0$.



        Edited: Instead of $C$ you can take any space $X$ with infinitely many path components. If you take $X = { 0 } cup { 1/n mid n in mathbb{N} }$ as in Eric Wofsey's answer, you get $Sigma X$ = Hawaiian earring.






        share|cite|improve this answer














        The Cantor set $C subset [0,1]$ has uncountably many path components (in fact, each single point subset is a path component of $C$). Hence $H_0(C)$ is a free abelian group with uncountably many generators.



        Taking the suspension $Sigma$ and noting that $tilde{H}_{n+1}(Sigma X) approx tilde{H}_n(X)$, where $tilde{H}_*$ denotes reduced homology, you can construct examples for all $H_i$ with $i ge 0$. Recall that $tilde{H}_i = H_i$ for $i > 0$.



        Edited: Instead of $C$ you can take any space $X$ with infinitely many path components. If you take $X = { 0 } cup { 1/n mid n in mathbb{N} }$ as in Eric Wofsey's answer, you get $Sigma X$ = Hawaiian earring.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 17 '18 at 15:39

























        answered Dec 2 '18 at 21:50









        Paul Frost

        9,2762631




        9,2762631






























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