What is wrong with this application of van Kampen's Theorem?












2












$begingroup$


My understanding of van Kampen's Theorem (simplified to just two neighbourhoods):



Let $X$ be a topological space and let ${N_a, N_b}$ be a cover of $X$ such that $N_a cap N_b$ is path-connected (and each open set is path-connected). Then,
$$
pi_1Xcong frac{pi_1N_a*pi_1N_b}{[i_a(gamma)][i_b(gamma)]^{-1}}
$$

Where $i_a$ and $i_b$ are the inclusion maps from $N_acap N_b$ to $N_a$ and $N_b$ respectively, and $gamma$ is any loop in the intersection $N_acap N_b$ (so we quotient by the normal subgroup generated by $[i_a(gamma)][i_b(gamma)]^{-1}$).



I was trying to make sure I understand it by seeing if I can "break" it: Let $N_a = N_b = X$ (where $X$ is a path-connected topological space). Then clearly $N_a$, $N_b$, and $N_acap N_b$ are path-connected and we may apply van Kampen. But now the intersection is just $X$, and so the inclusion maps $i_a$ and $i_b$ are just the identity maps, so applying van Kampen we have ($gamma$ in $X$)
$$
pi_1Xcong frac{pi_1N_a*pi_1N_b}{[gamma][gamma]^{-1}} cong pi_1X*pi_1X
$$

Clearly this is wrong, so I was wondering where my mistake is.










share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    My understanding of van Kampen's Theorem (simplified to just two neighbourhoods):



    Let $X$ be a topological space and let ${N_a, N_b}$ be a cover of $X$ such that $N_a cap N_b$ is path-connected (and each open set is path-connected). Then,
    $$
    pi_1Xcong frac{pi_1N_a*pi_1N_b}{[i_a(gamma)][i_b(gamma)]^{-1}}
    $$

    Where $i_a$ and $i_b$ are the inclusion maps from $N_acap N_b$ to $N_a$ and $N_b$ respectively, and $gamma$ is any loop in the intersection $N_acap N_b$ (so we quotient by the normal subgroup generated by $[i_a(gamma)][i_b(gamma)]^{-1}$).



    I was trying to make sure I understand it by seeing if I can "break" it: Let $N_a = N_b = X$ (where $X$ is a path-connected topological space). Then clearly $N_a$, $N_b$, and $N_acap N_b$ are path-connected and we may apply van Kampen. But now the intersection is just $X$, and so the inclusion maps $i_a$ and $i_b$ are just the identity maps, so applying van Kampen we have ($gamma$ in $X$)
    $$
    pi_1Xcong frac{pi_1N_a*pi_1N_b}{[gamma][gamma]^{-1}} cong pi_1X*pi_1X
    $$

    Clearly this is wrong, so I was wondering where my mistake is.










    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      My understanding of van Kampen's Theorem (simplified to just two neighbourhoods):



      Let $X$ be a topological space and let ${N_a, N_b}$ be a cover of $X$ such that $N_a cap N_b$ is path-connected (and each open set is path-connected). Then,
      $$
      pi_1Xcong frac{pi_1N_a*pi_1N_b}{[i_a(gamma)][i_b(gamma)]^{-1}}
      $$

      Where $i_a$ and $i_b$ are the inclusion maps from $N_acap N_b$ to $N_a$ and $N_b$ respectively, and $gamma$ is any loop in the intersection $N_acap N_b$ (so we quotient by the normal subgroup generated by $[i_a(gamma)][i_b(gamma)]^{-1}$).



      I was trying to make sure I understand it by seeing if I can "break" it: Let $N_a = N_b = X$ (where $X$ is a path-connected topological space). Then clearly $N_a$, $N_b$, and $N_acap N_b$ are path-connected and we may apply van Kampen. But now the intersection is just $X$, and so the inclusion maps $i_a$ and $i_b$ are just the identity maps, so applying van Kampen we have ($gamma$ in $X$)
      $$
      pi_1Xcong frac{pi_1N_a*pi_1N_b}{[gamma][gamma]^{-1}} cong pi_1X*pi_1X
      $$

      Clearly this is wrong, so I was wondering where my mistake is.










      share|cite|improve this question











      $endgroup$




      My understanding of van Kampen's Theorem (simplified to just two neighbourhoods):



      Let $X$ be a topological space and let ${N_a, N_b}$ be a cover of $X$ such that $N_a cap N_b$ is path-connected (and each open set is path-connected). Then,
      $$
      pi_1Xcong frac{pi_1N_a*pi_1N_b}{[i_a(gamma)][i_b(gamma)]^{-1}}
      $$

      Where $i_a$ and $i_b$ are the inclusion maps from $N_acap N_b$ to $N_a$ and $N_b$ respectively, and $gamma$ is any loop in the intersection $N_acap N_b$ (so we quotient by the normal subgroup generated by $[i_a(gamma)][i_b(gamma)]^{-1}$).



      I was trying to make sure I understand it by seeing if I can "break" it: Let $N_a = N_b = X$ (where $X$ is a path-connected topological space). Then clearly $N_a$, $N_b$, and $N_acap N_b$ are path-connected and we may apply van Kampen. But now the intersection is just $X$, and so the inclusion maps $i_a$ and $i_b$ are just the identity maps, so applying van Kampen we have ($gamma$ in $X$)
      $$
      pi_1Xcong frac{pi_1N_a*pi_1N_b}{[gamma][gamma]^{-1}} cong pi_1X*pi_1X
      $$

      Clearly this is wrong, so I was wondering where my mistake is.







      algebraic-topology fundamental-groups






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 12 '18 at 0:43









      Eric Wofsey

      185k14214341




      185k14214341










      asked Dec 12 '18 at 0:26









      FunctionalDefectFunctionalDefect

      255




      255






















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          Your mistake is in identifying the subgroup you are quotienting out. You quotient out by the normal subgroup generated by all elements of the form $[i_a(gamma)][i_b(gamma)]^{-1}$. When you write this, $[i_a(gamma)]$ is to be interpreted as an element of $pi_1(N_a)$, which is then considered as an element of $pi_1(N_a)*pi_1(N_b)$ via the canonical inclusion map $pi_1(N_a)to pi_1(N_a)*pi_1(N_b)$. Similarly, $[i_b(gamma)]$ is to be interpreted as an element of $pi_1(N_b)$, which is then considered as an element of $pi_1(N_a)*pi_1(N_b)$ via the canonical inclusion map $pi_1(N_b)to pi_1(N_a)*pi_1(N_b)$.



          What this means is that even in the case where $N_a=N_b=X$, $[i_a(gamma)]$ and $[i_b(gamma)]$ are not the same element of $pi_1(N_a)*pi_1(N_b)=pi_1(X)*pi_1(X)$. The first is the copy of $[gamma]$ in the first factor of $pi_1(X)*pi_1(X)$, and the second is the copy of $[gamma]$ in the second factor of $pi_1(X)*pi_1(X)$. So, modding out a relation that says these are equal amounts to identifying the two copies of $pi_1(X)$ inside $pi_1(X)*pi_1(X)$ with each other. When you do this, the result you get is just a single $pi_1(X)$ since the two copies have been identified, just like you want. (Of course, this is not a rigorous proof that the quotient is $pi_1(X)$, but it is the intuition from which you can build a proof.)






          share|cite|improve this answer









          $endgroup$













            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',
            autoActivateHeartbeat: false,
            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%2f3036058%2fwhat-is-wrong-with-this-application-of-van-kampens-theorem%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            Your mistake is in identifying the subgroup you are quotienting out. You quotient out by the normal subgroup generated by all elements of the form $[i_a(gamma)][i_b(gamma)]^{-1}$. When you write this, $[i_a(gamma)]$ is to be interpreted as an element of $pi_1(N_a)$, which is then considered as an element of $pi_1(N_a)*pi_1(N_b)$ via the canonical inclusion map $pi_1(N_a)to pi_1(N_a)*pi_1(N_b)$. Similarly, $[i_b(gamma)]$ is to be interpreted as an element of $pi_1(N_b)$, which is then considered as an element of $pi_1(N_a)*pi_1(N_b)$ via the canonical inclusion map $pi_1(N_b)to pi_1(N_a)*pi_1(N_b)$.



            What this means is that even in the case where $N_a=N_b=X$, $[i_a(gamma)]$ and $[i_b(gamma)]$ are not the same element of $pi_1(N_a)*pi_1(N_b)=pi_1(X)*pi_1(X)$. The first is the copy of $[gamma]$ in the first factor of $pi_1(X)*pi_1(X)$, and the second is the copy of $[gamma]$ in the second factor of $pi_1(X)*pi_1(X)$. So, modding out a relation that says these are equal amounts to identifying the two copies of $pi_1(X)$ inside $pi_1(X)*pi_1(X)$ with each other. When you do this, the result you get is just a single $pi_1(X)$ since the two copies have been identified, just like you want. (Of course, this is not a rigorous proof that the quotient is $pi_1(X)$, but it is the intuition from which you can build a proof.)






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              Your mistake is in identifying the subgroup you are quotienting out. You quotient out by the normal subgroup generated by all elements of the form $[i_a(gamma)][i_b(gamma)]^{-1}$. When you write this, $[i_a(gamma)]$ is to be interpreted as an element of $pi_1(N_a)$, which is then considered as an element of $pi_1(N_a)*pi_1(N_b)$ via the canonical inclusion map $pi_1(N_a)to pi_1(N_a)*pi_1(N_b)$. Similarly, $[i_b(gamma)]$ is to be interpreted as an element of $pi_1(N_b)$, which is then considered as an element of $pi_1(N_a)*pi_1(N_b)$ via the canonical inclusion map $pi_1(N_b)to pi_1(N_a)*pi_1(N_b)$.



              What this means is that even in the case where $N_a=N_b=X$, $[i_a(gamma)]$ and $[i_b(gamma)]$ are not the same element of $pi_1(N_a)*pi_1(N_b)=pi_1(X)*pi_1(X)$. The first is the copy of $[gamma]$ in the first factor of $pi_1(X)*pi_1(X)$, and the second is the copy of $[gamma]$ in the second factor of $pi_1(X)*pi_1(X)$. So, modding out a relation that says these are equal amounts to identifying the two copies of $pi_1(X)$ inside $pi_1(X)*pi_1(X)$ with each other. When you do this, the result you get is just a single $pi_1(X)$ since the two copies have been identified, just like you want. (Of course, this is not a rigorous proof that the quotient is $pi_1(X)$, but it is the intuition from which you can build a proof.)






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                Your mistake is in identifying the subgroup you are quotienting out. You quotient out by the normal subgroup generated by all elements of the form $[i_a(gamma)][i_b(gamma)]^{-1}$. When you write this, $[i_a(gamma)]$ is to be interpreted as an element of $pi_1(N_a)$, which is then considered as an element of $pi_1(N_a)*pi_1(N_b)$ via the canonical inclusion map $pi_1(N_a)to pi_1(N_a)*pi_1(N_b)$. Similarly, $[i_b(gamma)]$ is to be interpreted as an element of $pi_1(N_b)$, which is then considered as an element of $pi_1(N_a)*pi_1(N_b)$ via the canonical inclusion map $pi_1(N_b)to pi_1(N_a)*pi_1(N_b)$.



                What this means is that even in the case where $N_a=N_b=X$, $[i_a(gamma)]$ and $[i_b(gamma)]$ are not the same element of $pi_1(N_a)*pi_1(N_b)=pi_1(X)*pi_1(X)$. The first is the copy of $[gamma]$ in the first factor of $pi_1(X)*pi_1(X)$, and the second is the copy of $[gamma]$ in the second factor of $pi_1(X)*pi_1(X)$. So, modding out a relation that says these are equal amounts to identifying the two copies of $pi_1(X)$ inside $pi_1(X)*pi_1(X)$ with each other. When you do this, the result you get is just a single $pi_1(X)$ since the two copies have been identified, just like you want. (Of course, this is not a rigorous proof that the quotient is $pi_1(X)$, but it is the intuition from which you can build a proof.)






                share|cite|improve this answer









                $endgroup$



                Your mistake is in identifying the subgroup you are quotienting out. You quotient out by the normal subgroup generated by all elements of the form $[i_a(gamma)][i_b(gamma)]^{-1}$. When you write this, $[i_a(gamma)]$ is to be interpreted as an element of $pi_1(N_a)$, which is then considered as an element of $pi_1(N_a)*pi_1(N_b)$ via the canonical inclusion map $pi_1(N_a)to pi_1(N_a)*pi_1(N_b)$. Similarly, $[i_b(gamma)]$ is to be interpreted as an element of $pi_1(N_b)$, which is then considered as an element of $pi_1(N_a)*pi_1(N_b)$ via the canonical inclusion map $pi_1(N_b)to pi_1(N_a)*pi_1(N_b)$.



                What this means is that even in the case where $N_a=N_b=X$, $[i_a(gamma)]$ and $[i_b(gamma)]$ are not the same element of $pi_1(N_a)*pi_1(N_b)=pi_1(X)*pi_1(X)$. The first is the copy of $[gamma]$ in the first factor of $pi_1(X)*pi_1(X)$, and the second is the copy of $[gamma]$ in the second factor of $pi_1(X)*pi_1(X)$. So, modding out a relation that says these are equal amounts to identifying the two copies of $pi_1(X)$ inside $pi_1(X)*pi_1(X)$ with each other. When you do this, the result you get is just a single $pi_1(X)$ since the two copies have been identified, just like you want. (Of course, this is not a rigorous proof that the quotient is $pi_1(X)$, but it is the intuition from which you can build a proof.)







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 12 '18 at 0:43









                Eric WofseyEric Wofsey

                185k14214341




                185k14214341






























                    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.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3036058%2fwhat-is-wrong-with-this-application-of-van-kampens-theorem%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

                    Probability when a professor distributes a quiz and homework assignment to a class of n students.

                    Aardman Animations

                    Are they similar matrix