Can I use Seifert-van Kampen theorem infinite times The Next CEO of Stack OverflowPerturbation trick in the proof of Seifert-van-KampenSeifert-van-Kampen and free product with amalgamationFundamental group via Van KampenHow to use Seifert–van Kampen to find $pi_1( RP^n ∨ RP^n)$Practicing Seifert van KampenRealising Seifert-van Kampen in 2-complexesComputing fundamental group with Seifert-van Kampen theoremChoosing Open Sets in Applying the Seifert-Van Kampen Theorem to Topological space$H_1(F)$ for an orientable, compact, connected surfacehow to calculate the fundamental group using van kampen theorem?

Chain wire methods together in Lightning Web Components

Is it ever safe to open a suspicious HTML file (e.g. email attachment)?

How to place nodes around a circle from some initial angle?

Why don't programming languages automatically manage the synchronous/asynchronous problem?

What is the value of α and β in a triangle?

Where do students learn to solve polynomial equations these days?

I want to delete every two lines after 3rd lines in file contain very large number of lines :

Why do remote US companies require working in the US?

Why doesn't UK go for the same deal Japan has with EU to resolve Brexit?

Is there a difference between "Fahrstuhl" and "Aufzug"

Running a General Election and the European Elections together

How does Madhvacharya interpret Bhagavad Gita sloka 18.66?

Would a completely good Muggle be able to use a wand?

What is the difference between 翼 and 翅膀?

Display a text message if the shortcode is not found?

WOW air has ceased operation, can I get my tickets refunded?

Rotate a column

Is it my responsibility to learn a new technology in my own time my employer wants to implement?

Why the difference in type-inference over the as-pattern in two similar function definitions?

Won the lottery - how do I keep the money?

What happened in Rome, when the western empire "fell"?

Does soap repel water?

How did people program for Consoles with multiple CPUs?

Is it okay to majorly distort historical facts while writing a fiction story?



Can I use Seifert-van Kampen theorem infinite times



The Next CEO of Stack OverflowPerturbation trick in the proof of Seifert-van-KampenSeifert-van-Kampen and free product with amalgamationFundamental group via Van KampenHow to use Seifert–van Kampen to find $pi_1( RP^n ∨ RP^n)$Practicing Seifert van KampenRealising Seifert-van Kampen in 2-complexesComputing fundamental group with Seifert-van Kampen theoremChoosing Open Sets in Applying the Seifert-Van Kampen Theorem to Topological space$H_1(F)$ for an orientable, compact, connected surfacehow to calculate the fundamental group using van kampen theorem?










5












$begingroup$


I know the definition of Seifert-van Kampen theorem for a topological space "made" with 2 parts. Is not difficult to see that if I use the theorem a finite number of times to calculate a the fundamental group of a topological space made from finite many parts, it is valid. But, can I use the theorem infinite times? For example, to show that te fundamental group of a orientable surface of infinite genus is isomorphic to the free group with infinite generators?



Thanks.










share|cite|improve this question









$endgroup$
















    5












    $begingroup$


    I know the definition of Seifert-van Kampen theorem for a topological space "made" with 2 parts. Is not difficult to see that if I use the theorem a finite number of times to calculate a the fundamental group of a topological space made from finite many parts, it is valid. But, can I use the theorem infinite times? For example, to show that te fundamental group of a orientable surface of infinite genus is isomorphic to the free group with infinite generators?



    Thanks.










    share|cite|improve this question









    $endgroup$














      5












      5








      5


      1



      $begingroup$


      I know the definition of Seifert-van Kampen theorem for a topological space "made" with 2 parts. Is not difficult to see that if I use the theorem a finite number of times to calculate a the fundamental group of a topological space made from finite many parts, it is valid. But, can I use the theorem infinite times? For example, to show that te fundamental group of a orientable surface of infinite genus is isomorphic to the free group with infinite generators?



      Thanks.










      share|cite|improve this question









      $endgroup$




      I know the definition of Seifert-van Kampen theorem for a topological space "made" with 2 parts. Is not difficult to see that if I use the theorem a finite number of times to calculate a the fundamental group of a topological space made from finite many parts, it is valid. But, can I use the theorem infinite times? For example, to show that te fundamental group of a orientable surface of infinite genus is isomorphic to the free group with infinite generators?



      Thanks.







      general-topology algebraic-topology fundamental-groups






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 24 at 4:48









      J.RodriguezJ.Rodriguez

      19710




      19710




















          3 Answers
          3






          active

          oldest

          votes


















          5












          $begingroup$

          If we have an infinite cover of a space $X$, one approach is to compute $pi_1$ of finite unions of those covers, thus reducing the problem to the case of an increasing union $U_1 subseteq U_2 subseteq cdots subseteq X = bigcup_i=1^infty U_i$. This is handled via a direct limit of the fundamental groups of the open sets $U_i$.




          Given a space $X$ which is an increasing union of open sets $U_1 subseteq U_2 subseteq ldots$, with a specified basepoint $ast in U_1$, we have a diagram
          $$requireAMScd
          beginCD
          pi_1(U_1,ast) @> >> pi_1(U_2,ast) @>>> ldots \
          @VVV @VVV \
          pi_1(X, ast) @= pi_1(X,ast) @= ldots
          endCD$$

          which gives a map from the direct limit
          $$Phi: varinjlim pi_1(U_i, ast) to pi_1(X,ast).$$




          Claim: $Phi: varinjlim pi_1(U_i, ast) to pi_1(X,ast)$ is an isomorphism.




          Proof: To see that $Phi$ is surjective, observe that if $gamma: (S^1, ast) to (X,ast)$ is a based loop, then $gamma(S^1) subseteq cup_i=1^infty U_i$ is compact, so its image is contained in a finite subcover of $U_i_i=1^infty$ and thus in $U_i$ for some $i$.



          In the wise words of one of my topology professors, "injectivity is just surjectivity, one dimension higher." If $[gamma_1]$ and $[gamma_2]$ are homotopy classes in the direct limit which are (based) homotopic in $X$ via $H: S^1 times I to X$, then the image of $H$ is (by the same argument) contained in $U_j$ for some $j$, which insures that $[gamma_1] = [gamma_2]$ in $pi_1(U_k,ast)$ for $k geq j$. This shows injectivity.




          While a direct limit of a system of groups may be intimidating, it allows us to compute. For instance, if $F_n$ is the free group on generators $x_1,x_2,ldots, x_n$, with maps $F_n to F_n+1$ given by including the generators $x_1,x_2,ldots, x_n subseteq x_1,x_2,ldots, x_n+1$, then
          $$ varinjlim F_n = F_infty,$$
          where $F_infty$ is the free group on infinitely many generators $x_1, x_2,ldots$.






          share|cite|improve this answer









          $endgroup$




















            4












            $begingroup$

            As Joshua Mundinger has nicely explained, you can "iterate" the Seifert-van Kampen theorem infinitely many times by taking a direct limit, using the fact that $pi_1$ preserves direct limits of open inclusions. However, there is also a more direct way to use Seifert-van Kampen for infinite open covers: there is a version of the theorem that applies to covers by an arbitrary number of open sets, rather than just two open sets.



            Here's one version of the statement (this is Theorem 1.20 in Hatcher's Algebraic Topology, for instance). Suppose $(X,*)$ is a pointed space and $(U_i)_iin I$ is an open cover of $X$ such that $*in U_i$ for all $i$. Suppose furthermore that $U_i$, $U_icap U_j$, and $U_icap U_jcap U_k$ are path-connected for all $i,j,kin I$. Then $pi_1(X,*)$ is isomorphic (via the obvious map) to the quotient of the free product of the groups $pi_1(U_i,*)$ by relations which say the two maps $pi_1(U_icap U_j,*)to pi_1(U_i,*)$ and $pi_1(U_icap U_j,*)topi_1(U_j,*)$ become equal.



            Note that this is useful not just for infinite covers, but also for finite covers by more than two sets, allowing you to compute the final result all at once rather than needing to iterate.






            share|cite|improve this answer









            $endgroup$




















              0












              $begingroup$

              What would it even mean to apply the theorem infinitely many times? That would have to be some sort of inductive process - and you have to be very careful about the sorts of infinities you get out of that.



              In this case, adding one hole at a time, we get as subgroups a chain of free groups on increasing numbers of generators - but since there are still infinitely many holes at any stage of the process, there will be another factor that we simply can't cleanly measure this way.



              Now, we can take the direct limit of those free subgroups we found to say that there's an infinite free group that's a subgroup of the fundamental group. We can't say more without additional tools.






              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%2f3160109%2fcan-i-use-seifert-van-kampen-theorem-infinite-times%23new-answer', 'question_page');

                );

                Post as a guest















                Required, but never shown

























                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                5












                $begingroup$

                If we have an infinite cover of a space $X$, one approach is to compute $pi_1$ of finite unions of those covers, thus reducing the problem to the case of an increasing union $U_1 subseteq U_2 subseteq cdots subseteq X = bigcup_i=1^infty U_i$. This is handled via a direct limit of the fundamental groups of the open sets $U_i$.




                Given a space $X$ which is an increasing union of open sets $U_1 subseteq U_2 subseteq ldots$, with a specified basepoint $ast in U_1$, we have a diagram
                $$requireAMScd
                beginCD
                pi_1(U_1,ast) @> >> pi_1(U_2,ast) @>>> ldots \
                @VVV @VVV \
                pi_1(X, ast) @= pi_1(X,ast) @= ldots
                endCD$$

                which gives a map from the direct limit
                $$Phi: varinjlim pi_1(U_i, ast) to pi_1(X,ast).$$




                Claim: $Phi: varinjlim pi_1(U_i, ast) to pi_1(X,ast)$ is an isomorphism.




                Proof: To see that $Phi$ is surjective, observe that if $gamma: (S^1, ast) to (X,ast)$ is a based loop, then $gamma(S^1) subseteq cup_i=1^infty U_i$ is compact, so its image is contained in a finite subcover of $U_i_i=1^infty$ and thus in $U_i$ for some $i$.



                In the wise words of one of my topology professors, "injectivity is just surjectivity, one dimension higher." If $[gamma_1]$ and $[gamma_2]$ are homotopy classes in the direct limit which are (based) homotopic in $X$ via $H: S^1 times I to X$, then the image of $H$ is (by the same argument) contained in $U_j$ for some $j$, which insures that $[gamma_1] = [gamma_2]$ in $pi_1(U_k,ast)$ for $k geq j$. This shows injectivity.




                While a direct limit of a system of groups may be intimidating, it allows us to compute. For instance, if $F_n$ is the free group on generators $x_1,x_2,ldots, x_n$, with maps $F_n to F_n+1$ given by including the generators $x_1,x_2,ldots, x_n subseteq x_1,x_2,ldots, x_n+1$, then
                $$ varinjlim F_n = F_infty,$$
                where $F_infty$ is the free group on infinitely many generators $x_1, x_2,ldots$.






                share|cite|improve this answer









                $endgroup$

















                  5












                  $begingroup$

                  If we have an infinite cover of a space $X$, one approach is to compute $pi_1$ of finite unions of those covers, thus reducing the problem to the case of an increasing union $U_1 subseteq U_2 subseteq cdots subseteq X = bigcup_i=1^infty U_i$. This is handled via a direct limit of the fundamental groups of the open sets $U_i$.




                  Given a space $X$ which is an increasing union of open sets $U_1 subseteq U_2 subseteq ldots$, with a specified basepoint $ast in U_1$, we have a diagram
                  $$requireAMScd
                  beginCD
                  pi_1(U_1,ast) @> >> pi_1(U_2,ast) @>>> ldots \
                  @VVV @VVV \
                  pi_1(X, ast) @= pi_1(X,ast) @= ldots
                  endCD$$

                  which gives a map from the direct limit
                  $$Phi: varinjlim pi_1(U_i, ast) to pi_1(X,ast).$$




                  Claim: $Phi: varinjlim pi_1(U_i, ast) to pi_1(X,ast)$ is an isomorphism.




                  Proof: To see that $Phi$ is surjective, observe that if $gamma: (S^1, ast) to (X,ast)$ is a based loop, then $gamma(S^1) subseteq cup_i=1^infty U_i$ is compact, so its image is contained in a finite subcover of $U_i_i=1^infty$ and thus in $U_i$ for some $i$.



                  In the wise words of one of my topology professors, "injectivity is just surjectivity, one dimension higher." If $[gamma_1]$ and $[gamma_2]$ are homotopy classes in the direct limit which are (based) homotopic in $X$ via $H: S^1 times I to X$, then the image of $H$ is (by the same argument) contained in $U_j$ for some $j$, which insures that $[gamma_1] = [gamma_2]$ in $pi_1(U_k,ast)$ for $k geq j$. This shows injectivity.




                  While a direct limit of a system of groups may be intimidating, it allows us to compute. For instance, if $F_n$ is the free group on generators $x_1,x_2,ldots, x_n$, with maps $F_n to F_n+1$ given by including the generators $x_1,x_2,ldots, x_n subseteq x_1,x_2,ldots, x_n+1$, then
                  $$ varinjlim F_n = F_infty,$$
                  where $F_infty$ is the free group on infinitely many generators $x_1, x_2,ldots$.






                  share|cite|improve this answer









                  $endgroup$















                    5












                    5








                    5





                    $begingroup$

                    If we have an infinite cover of a space $X$, one approach is to compute $pi_1$ of finite unions of those covers, thus reducing the problem to the case of an increasing union $U_1 subseteq U_2 subseteq cdots subseteq X = bigcup_i=1^infty U_i$. This is handled via a direct limit of the fundamental groups of the open sets $U_i$.




                    Given a space $X$ which is an increasing union of open sets $U_1 subseteq U_2 subseteq ldots$, with a specified basepoint $ast in U_1$, we have a diagram
                    $$requireAMScd
                    beginCD
                    pi_1(U_1,ast) @> >> pi_1(U_2,ast) @>>> ldots \
                    @VVV @VVV \
                    pi_1(X, ast) @= pi_1(X,ast) @= ldots
                    endCD$$

                    which gives a map from the direct limit
                    $$Phi: varinjlim pi_1(U_i, ast) to pi_1(X,ast).$$




                    Claim: $Phi: varinjlim pi_1(U_i, ast) to pi_1(X,ast)$ is an isomorphism.




                    Proof: To see that $Phi$ is surjective, observe that if $gamma: (S^1, ast) to (X,ast)$ is a based loop, then $gamma(S^1) subseteq cup_i=1^infty U_i$ is compact, so its image is contained in a finite subcover of $U_i_i=1^infty$ and thus in $U_i$ for some $i$.



                    In the wise words of one of my topology professors, "injectivity is just surjectivity, one dimension higher." If $[gamma_1]$ and $[gamma_2]$ are homotopy classes in the direct limit which are (based) homotopic in $X$ via $H: S^1 times I to X$, then the image of $H$ is (by the same argument) contained in $U_j$ for some $j$, which insures that $[gamma_1] = [gamma_2]$ in $pi_1(U_k,ast)$ for $k geq j$. This shows injectivity.




                    While a direct limit of a system of groups may be intimidating, it allows us to compute. For instance, if $F_n$ is the free group on generators $x_1,x_2,ldots, x_n$, with maps $F_n to F_n+1$ given by including the generators $x_1,x_2,ldots, x_n subseteq x_1,x_2,ldots, x_n+1$, then
                    $$ varinjlim F_n = F_infty,$$
                    where $F_infty$ is the free group on infinitely many generators $x_1, x_2,ldots$.






                    share|cite|improve this answer









                    $endgroup$



                    If we have an infinite cover of a space $X$, one approach is to compute $pi_1$ of finite unions of those covers, thus reducing the problem to the case of an increasing union $U_1 subseteq U_2 subseteq cdots subseteq X = bigcup_i=1^infty U_i$. This is handled via a direct limit of the fundamental groups of the open sets $U_i$.




                    Given a space $X$ which is an increasing union of open sets $U_1 subseteq U_2 subseteq ldots$, with a specified basepoint $ast in U_1$, we have a diagram
                    $$requireAMScd
                    beginCD
                    pi_1(U_1,ast) @> >> pi_1(U_2,ast) @>>> ldots \
                    @VVV @VVV \
                    pi_1(X, ast) @= pi_1(X,ast) @= ldots
                    endCD$$

                    which gives a map from the direct limit
                    $$Phi: varinjlim pi_1(U_i, ast) to pi_1(X,ast).$$




                    Claim: $Phi: varinjlim pi_1(U_i, ast) to pi_1(X,ast)$ is an isomorphism.




                    Proof: To see that $Phi$ is surjective, observe that if $gamma: (S^1, ast) to (X,ast)$ is a based loop, then $gamma(S^1) subseteq cup_i=1^infty U_i$ is compact, so its image is contained in a finite subcover of $U_i_i=1^infty$ and thus in $U_i$ for some $i$.



                    In the wise words of one of my topology professors, "injectivity is just surjectivity, one dimension higher." If $[gamma_1]$ and $[gamma_2]$ are homotopy classes in the direct limit which are (based) homotopic in $X$ via $H: S^1 times I to X$, then the image of $H$ is (by the same argument) contained in $U_j$ for some $j$, which insures that $[gamma_1] = [gamma_2]$ in $pi_1(U_k,ast)$ for $k geq j$. This shows injectivity.




                    While a direct limit of a system of groups may be intimidating, it allows us to compute. For instance, if $F_n$ is the free group on generators $x_1,x_2,ldots, x_n$, with maps $F_n to F_n+1$ given by including the generators $x_1,x_2,ldots, x_n subseteq x_1,x_2,ldots, x_n+1$, then
                    $$ varinjlim F_n = F_infty,$$
                    where $F_infty$ is the free group on infinitely many generators $x_1, x_2,ldots$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Mar 24 at 5:31









                    Joshua MundingerJoshua Mundinger

                    2,9071028




                    2,9071028





















                        4












                        $begingroup$

                        As Joshua Mundinger has nicely explained, you can "iterate" the Seifert-van Kampen theorem infinitely many times by taking a direct limit, using the fact that $pi_1$ preserves direct limits of open inclusions. However, there is also a more direct way to use Seifert-van Kampen for infinite open covers: there is a version of the theorem that applies to covers by an arbitrary number of open sets, rather than just two open sets.



                        Here's one version of the statement (this is Theorem 1.20 in Hatcher's Algebraic Topology, for instance). Suppose $(X,*)$ is a pointed space and $(U_i)_iin I$ is an open cover of $X$ such that $*in U_i$ for all $i$. Suppose furthermore that $U_i$, $U_icap U_j$, and $U_icap U_jcap U_k$ are path-connected for all $i,j,kin I$. Then $pi_1(X,*)$ is isomorphic (via the obvious map) to the quotient of the free product of the groups $pi_1(U_i,*)$ by relations which say the two maps $pi_1(U_icap U_j,*)to pi_1(U_i,*)$ and $pi_1(U_icap U_j,*)topi_1(U_j,*)$ become equal.



                        Note that this is useful not just for infinite covers, but also for finite covers by more than two sets, allowing you to compute the final result all at once rather than needing to iterate.






                        share|cite|improve this answer









                        $endgroup$

















                          4












                          $begingroup$

                          As Joshua Mundinger has nicely explained, you can "iterate" the Seifert-van Kampen theorem infinitely many times by taking a direct limit, using the fact that $pi_1$ preserves direct limits of open inclusions. However, there is also a more direct way to use Seifert-van Kampen for infinite open covers: there is a version of the theorem that applies to covers by an arbitrary number of open sets, rather than just two open sets.



                          Here's one version of the statement (this is Theorem 1.20 in Hatcher's Algebraic Topology, for instance). Suppose $(X,*)$ is a pointed space and $(U_i)_iin I$ is an open cover of $X$ such that $*in U_i$ for all $i$. Suppose furthermore that $U_i$, $U_icap U_j$, and $U_icap U_jcap U_k$ are path-connected for all $i,j,kin I$. Then $pi_1(X,*)$ is isomorphic (via the obvious map) to the quotient of the free product of the groups $pi_1(U_i,*)$ by relations which say the two maps $pi_1(U_icap U_j,*)to pi_1(U_i,*)$ and $pi_1(U_icap U_j,*)topi_1(U_j,*)$ become equal.



                          Note that this is useful not just for infinite covers, but also for finite covers by more than two sets, allowing you to compute the final result all at once rather than needing to iterate.






                          share|cite|improve this answer









                          $endgroup$















                            4












                            4








                            4





                            $begingroup$

                            As Joshua Mundinger has nicely explained, you can "iterate" the Seifert-van Kampen theorem infinitely many times by taking a direct limit, using the fact that $pi_1$ preserves direct limits of open inclusions. However, there is also a more direct way to use Seifert-van Kampen for infinite open covers: there is a version of the theorem that applies to covers by an arbitrary number of open sets, rather than just two open sets.



                            Here's one version of the statement (this is Theorem 1.20 in Hatcher's Algebraic Topology, for instance). Suppose $(X,*)$ is a pointed space and $(U_i)_iin I$ is an open cover of $X$ such that $*in U_i$ for all $i$. Suppose furthermore that $U_i$, $U_icap U_j$, and $U_icap U_jcap U_k$ are path-connected for all $i,j,kin I$. Then $pi_1(X,*)$ is isomorphic (via the obvious map) to the quotient of the free product of the groups $pi_1(U_i,*)$ by relations which say the two maps $pi_1(U_icap U_j,*)to pi_1(U_i,*)$ and $pi_1(U_icap U_j,*)topi_1(U_j,*)$ become equal.



                            Note that this is useful not just for infinite covers, but also for finite covers by more than two sets, allowing you to compute the final result all at once rather than needing to iterate.






                            share|cite|improve this answer









                            $endgroup$



                            As Joshua Mundinger has nicely explained, you can "iterate" the Seifert-van Kampen theorem infinitely many times by taking a direct limit, using the fact that $pi_1$ preserves direct limits of open inclusions. However, there is also a more direct way to use Seifert-van Kampen for infinite open covers: there is a version of the theorem that applies to covers by an arbitrary number of open sets, rather than just two open sets.



                            Here's one version of the statement (this is Theorem 1.20 in Hatcher's Algebraic Topology, for instance). Suppose $(X,*)$ is a pointed space and $(U_i)_iin I$ is an open cover of $X$ such that $*in U_i$ for all $i$. Suppose furthermore that $U_i$, $U_icap U_j$, and $U_icap U_jcap U_k$ are path-connected for all $i,j,kin I$. Then $pi_1(X,*)$ is isomorphic (via the obvious map) to the quotient of the free product of the groups $pi_1(U_i,*)$ by relations which say the two maps $pi_1(U_icap U_j,*)to pi_1(U_i,*)$ and $pi_1(U_icap U_j,*)topi_1(U_j,*)$ become equal.



                            Note that this is useful not just for infinite covers, but also for finite covers by more than two sets, allowing you to compute the final result all at once rather than needing to iterate.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Mar 24 at 6:22









                            Eric WofseyEric Wofsey

                            191k14216349




                            191k14216349





















                                0












                                $begingroup$

                                What would it even mean to apply the theorem infinitely many times? That would have to be some sort of inductive process - and you have to be very careful about the sorts of infinities you get out of that.



                                In this case, adding one hole at a time, we get as subgroups a chain of free groups on increasing numbers of generators - but since there are still infinitely many holes at any stage of the process, there will be another factor that we simply can't cleanly measure this way.



                                Now, we can take the direct limit of those free subgroups we found to say that there's an infinite free group that's a subgroup of the fundamental group. We can't say more without additional tools.






                                share|cite|improve this answer









                                $endgroup$

















                                  0












                                  $begingroup$

                                  What would it even mean to apply the theorem infinitely many times? That would have to be some sort of inductive process - and you have to be very careful about the sorts of infinities you get out of that.



                                  In this case, adding one hole at a time, we get as subgroups a chain of free groups on increasing numbers of generators - but since there are still infinitely many holes at any stage of the process, there will be another factor that we simply can't cleanly measure this way.



                                  Now, we can take the direct limit of those free subgroups we found to say that there's an infinite free group that's a subgroup of the fundamental group. We can't say more without additional tools.






                                  share|cite|improve this answer









                                  $endgroup$















                                    0












                                    0








                                    0





                                    $begingroup$

                                    What would it even mean to apply the theorem infinitely many times? That would have to be some sort of inductive process - and you have to be very careful about the sorts of infinities you get out of that.



                                    In this case, adding one hole at a time, we get as subgroups a chain of free groups on increasing numbers of generators - but since there are still infinitely many holes at any stage of the process, there will be another factor that we simply can't cleanly measure this way.



                                    Now, we can take the direct limit of those free subgroups we found to say that there's an infinite free group that's a subgroup of the fundamental group. We can't say more without additional tools.






                                    share|cite|improve this answer









                                    $endgroup$



                                    What would it even mean to apply the theorem infinitely many times? That would have to be some sort of inductive process - and you have to be very careful about the sorts of infinities you get out of that.



                                    In this case, adding one hole at a time, we get as subgroups a chain of free groups on increasing numbers of generators - but since there are still infinitely many holes at any stage of the process, there will be another factor that we simply can't cleanly measure this way.



                                    Now, we can take the direct limit of those free subgroups we found to say that there's an infinite free group that's a subgroup of the fundamental group. We can't say more without additional tools.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Mar 24 at 5:27









                                    jmerryjmerry

                                    16.9k11633




                                    16.9k11633



























                                        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%2f3160109%2fcan-i-use-seifert-van-kampen-theorem-infinite-times%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

                                        Adding axes to figuresAdding axes labels to LaTeX figuresLaTeX equivalent of ConTeXt buffersRotate a node but not its content: the case of the ellipse decorationHow to define the default vertical distance between nodes?TikZ scaling graphic and adjust node position and keep font sizeNumerical conditional within tikz keys?adding axes to shapesAlign axes across subfiguresAdding figures with a certain orderLine up nested tikz enviroments or how to get rid of themAdding axes labels to LaTeX figures

                                        Tähtien Talli Jäsenet | Lähteet | NavigointivalikkoSuomen Hippos – Tähtien Talli

                                        Do these cracks on my tires look bad? The Next CEO of Stack OverflowDry rot tire should I replace?Having to replace tiresFishtailed so easily? Bad tires? ABS?Filling the tires with something other than air, to avoid puncture hassles?Used Michelin tires safe to install?Do these tyre cracks necessitate replacement?Rumbling noise: tires or mechanicalIs it possible to fix noisy feathered tires?Are bad winter tires still better than summer tires in winter?Torque converter failure - Related to replacing only 2 tires?Why use snow tires on all 4 wheels on 2-wheel-drive cars?