Construct compact space with some homology group not finitely generated











up vote
2
down vote

favorite












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.










share|cite|improve this question


















  • 1




    This thread might interest you: math.stackexchange.com/questions/2512580/…
    – Aleksandar Milivojevic
    4 hours ago















up vote
2
down vote

favorite












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.










share|cite|improve this question


















  • 1




    This thread might interest you: math.stackexchange.com/questions/2512580/…
    – Aleksandar Milivojevic
    4 hours ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite











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.










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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 5 hours ago









CHwC

229111




229111








  • 1




    This thread might interest you: math.stackexchange.com/questions/2512580/…
    – Aleksandar Milivojevic
    4 hours ago














  • 1




    This thread might interest you: math.stackexchange.com/questions/2512580/…
    – Aleksandar Milivojevic
    4 hours ago








1




1




This thread might interest you: math.stackexchange.com/questions/2512580/…
– Aleksandar Milivojevic
4 hours ago




This thread might interest you: math.stackexchange.com/questions/2512580/…
– Aleksandar Milivojevic
4 hours ago










2 Answers
2






active

oldest

votes

















up vote
3
down vote













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
    4 hours ago










  • okay the second point is simply the definition of singular homology, but what about the first?
    – CHwC
    3 hours ago










  • 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
    2 hours ago


















up vote
3
down vote













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 $i > 0$.






share|cite|improve this answer























    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023089%2fconstruct-compact-space-with-some-homology-group-not-finitely-generated%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote













    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
      4 hours ago










    • okay the second point is simply the definition of singular homology, but what about the first?
      – CHwC
      3 hours ago










    • 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
      2 hours ago















    up vote
    3
    down vote













    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
      4 hours ago










    • okay the second point is simply the definition of singular homology, but what about the first?
      – CHwC
      3 hours ago










    • 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
      2 hours ago













    up vote
    3
    down vote










    up vote
    3
    down vote









    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 4 hours ago









    Eric Wofsey

    176k12202326




    176k12202326












    • 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
      4 hours ago










    • okay the second point is simply the definition of singular homology, but what about the first?
      – CHwC
      3 hours ago










    • 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
      2 hours ago


















    • 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
      4 hours ago










    • okay the second point is simply the definition of singular homology, but what about the first?
      – CHwC
      3 hours ago










    • 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
      2 hours ago
















    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
    4 hours ago




    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
    4 hours ago












    okay the second point is simply the definition of singular homology, but what about the first?
    – CHwC
    3 hours ago




    okay the second point is simply the definition of singular homology, but what about the first?
    – CHwC
    3 hours ago












    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
    2 hours ago




    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
    2 hours ago










    up vote
    3
    down vote













    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 $i > 0$.






    share|cite|improve this answer



























      up vote
      3
      down vote













      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 $i > 0$.






      share|cite|improve this answer

























        up vote
        3
        down vote










        up vote
        3
        down vote









        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 $i > 0$.






        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 $i > 0$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 1 hour ago

























        answered 3 hours ago









        Paul Frost

        7,8841527




        7,8841527






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023089%2fconstruct-compact-space-with-some-homology-group-not-finitely-generated%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            U.S. state

            Michael Jordan

            Prague