Is there a non trivial covering of the Klein bottle by the Klein bottle












1












$begingroup$


Let K be the Klein bottle obtained by the quotient of $[0, 1] × [0; 1]$
by the equivalence relation $(x, 0) ∼ (1 − x, 1)$ and $(0, y) ∼ (1, y)$.



Is there a non trivial covering of $K$ by $K$?



The universal cover of $K$ is $Bbb R^2$ and I know the torus can also be a cover of $K$, but I don't know where to start.



Thank you for any hints and help.






general-topology algebraic-topology klein-bottle







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let K be the Klein bottle obtained by the quotient of $[0, 1] × [0; 1]$
    by the equivalence relation $(x, 0) ∼ (1 − x, 1)$ and $(0, y) ∼ (1, y)$.



    Is there a non trivial covering of $K$ by $K$?



    The universal cover of $K$ is $Bbb R^2$ and I know the torus can also be a cover of $K$, but I don't know where to start.



    Thank you for any hints and help.






    general-topology algebraic-topology klein-bottle







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let K be the Klein bottle obtained by the quotient of $[0, 1] × [0; 1]$
      by the equivalence relation $(x, 0) ∼ (1 − x, 1)$ and $(0, y) ∼ (1, y)$.



      Is there a non trivial covering of $K$ by $K$?



      The universal cover of $K$ is $Bbb R^2$ and I know the torus can also be a cover of $K$, but I don't know where to start.



      Thank you for any hints and help.






      general-topology algebraic-topology klein-bottle







      share|cite|improve this question









      $endgroup$




      Let K be the Klein bottle obtained by the quotient of $[0, 1] × [0; 1]$
      by the equivalence relation $(x, 0) ∼ (1 − x, 1)$ and $(0, y) ∼ (1, y)$.



      Is there a non trivial covering of $K$ by $K$?



      The universal cover of $K$ is $Bbb R^2$ and I know the torus can also be a cover of $K$, but I don't know where to start.



      Thank you for any hints and help.






      general-topology algebraic-topology klein-bottle




      general-topology algebraic-topology klein-bottle






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 2 hours ago









      PerelManPerelMan

      629312




      629312






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          The Klein bottle is the quotient of $mathbb{R}^2$ by the group $G$ generated by $u(x,y)=(1-x,y)$ and $v(x,y)=(x,y+1)$Consider $f(x,y)=(x,2y)$ $fcirc u(x,y)=f(1-x,y)=(1-x,2y)=ucirc f(x,y)$.



          $fcirc v(x,y)=f(x,y+1)=(x,2y+2)=v^2circ f$. This implies that $f$ induces a continuous map of $mathbb{R}^2/G$ this map is a covering of order $2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            this is a cover of the Klein bottle by the Torus right? or it is a cover by the Klein bottle?
            $endgroup$
            – PerelMan
            1 hour ago





















          2












          $begingroup$

          One way you can envision the two-fold cover of $K$ by the torus by placing two copies of the given square next to each other such that the $(x,0)$ side of one is touching the $(x,1)$ side of the other. To check that this translates to a well-defined map $Tto K$ is fairly straightforward.



          This can be extended to a 3-fold cover of $K$ by itself if you place three such squares next to each other (or more generally for any odd $n$).






          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%2f3125520%2fis-there-a-non-trivial-covering-of-the-klein-bottle-by-the-klein-bottle%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









            3












            $begingroup$

            The Klein bottle is the quotient of $mathbb{R}^2$ by the group $G$ generated by $u(x,y)=(1-x,y)$ and $v(x,y)=(x,y+1)$Consider $f(x,y)=(x,2y)$ $fcirc u(x,y)=f(1-x,y)=(1-x,2y)=ucirc f(x,y)$.



            $fcirc v(x,y)=f(x,y+1)=(x,2y+2)=v^2circ f$. This implies that $f$ induces a continuous map of $mathbb{R}^2/G$ this map is a covering of order $2$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              this is a cover of the Klein bottle by the Torus right? or it is a cover by the Klein bottle?
              $endgroup$
              – PerelMan
              1 hour ago


















            3












            $begingroup$

            The Klein bottle is the quotient of $mathbb{R}^2$ by the group $G$ generated by $u(x,y)=(1-x,y)$ and $v(x,y)=(x,y+1)$Consider $f(x,y)=(x,2y)$ $fcirc u(x,y)=f(1-x,y)=(1-x,2y)=ucirc f(x,y)$.



            $fcirc v(x,y)=f(x,y+1)=(x,2y+2)=v^2circ f$. This implies that $f$ induces a continuous map of $mathbb{R}^2/G$ this map is a covering of order $2$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              this is a cover of the Klein bottle by the Torus right? or it is a cover by the Klein bottle?
              $endgroup$
              – PerelMan
              1 hour ago
















            3












            3








            3





            $begingroup$

            The Klein bottle is the quotient of $mathbb{R}^2$ by the group $G$ generated by $u(x,y)=(1-x,y)$ and $v(x,y)=(x,y+1)$Consider $f(x,y)=(x,2y)$ $fcirc u(x,y)=f(1-x,y)=(1-x,2y)=ucirc f(x,y)$.



            $fcirc v(x,y)=f(x,y+1)=(x,2y+2)=v^2circ f$. This implies that $f$ induces a continuous map of $mathbb{R}^2/G$ this map is a covering of order $2$.






            share|cite|improve this answer









            $endgroup$



            The Klein bottle is the quotient of $mathbb{R}^2$ by the group $G$ generated by $u(x,y)=(1-x,y)$ and $v(x,y)=(x,y+1)$Consider $f(x,y)=(x,2y)$ $fcirc u(x,y)=f(1-x,y)=(1-x,2y)=ucirc f(x,y)$.



            $fcirc v(x,y)=f(x,y+1)=(x,2y+2)=v^2circ f$. This implies that $f$ induces a continuous map of $mathbb{R}^2/G$ this map is a covering of order $2$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 hours ago









            Tsemo AristideTsemo Aristide

            58.7k11445




            58.7k11445












            • $begingroup$
              this is a cover of the Klein bottle by the Torus right? or it is a cover by the Klein bottle?
              $endgroup$
              – PerelMan
              1 hour ago




















            • $begingroup$
              this is a cover of the Klein bottle by the Torus right? or it is a cover by the Klein bottle?
              $endgroup$
              – PerelMan
              1 hour ago


















            $begingroup$
            this is a cover of the Klein bottle by the Torus right? or it is a cover by the Klein bottle?
            $endgroup$
            – PerelMan
            1 hour ago






            $begingroup$
            this is a cover of the Klein bottle by the Torus right? or it is a cover by the Klein bottle?
            $endgroup$
            – PerelMan
            1 hour ago













            2












            $begingroup$

            One way you can envision the two-fold cover of $K$ by the torus by placing two copies of the given square next to each other such that the $(x,0)$ side of one is touching the $(x,1)$ side of the other. To check that this translates to a well-defined map $Tto K$ is fairly straightforward.



            This can be extended to a 3-fold cover of $K$ by itself if you place three such squares next to each other (or more generally for any odd $n$).






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              One way you can envision the two-fold cover of $K$ by the torus by placing two copies of the given square next to each other such that the $(x,0)$ side of one is touching the $(x,1)$ side of the other. To check that this translates to a well-defined map $Tto K$ is fairly straightforward.



              This can be extended to a 3-fold cover of $K$ by itself if you place three such squares next to each other (or more generally for any odd $n$).






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                One way you can envision the two-fold cover of $K$ by the torus by placing two copies of the given square next to each other such that the $(x,0)$ side of one is touching the $(x,1)$ side of the other. To check that this translates to a well-defined map $Tto K$ is fairly straightforward.



                This can be extended to a 3-fold cover of $K$ by itself if you place three such squares next to each other (or more generally for any odd $n$).






                share|cite|improve this answer









                $endgroup$



                One way you can envision the two-fold cover of $K$ by the torus by placing two copies of the given square next to each other such that the $(x,0)$ side of one is touching the $(x,1)$ side of the other. To check that this translates to a well-defined map $Tto K$ is fairly straightforward.



                This can be extended to a 3-fold cover of $K$ by itself if you place three such squares next to each other (or more generally for any odd $n$).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 hours ago









                Rolf HoyerRolf Hoyer

                11.2k31629




                11.2k31629






























                    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%2f3125520%2fis-there-a-non-trivial-covering-of-the-klein-bottle-by-the-klein-bottle%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

                    Ponta tanko

                    Image
                    Usona ponta tanko M60A1 Ponta tanko estas, simile kiel kirasita savveturilo, inter specialaj kirasitaj transportiloj. Kiel pontaj tankoj eblas konsideri ĉiujn memmovigaj rimedoj ebligantaj superi obstaklojn ĝis larĝeco 24 metroj rekte en spaco de bataloperacoj. Ties ĉefa tasko estas ebligi al la propraj taĉmentoj daŭrigi en rapida antaŭenirado sen senbezona reteniĝo. El tio pli-malpli rezultas taktikaj-teknikaj postuloj, kiujn ili devas plenumi. Precipe temas pri kapablo operacii komune kun unue-sinsekvaj taĉmentoj kaj transpontigi obstaklojn dum kelke da minutoj. This page is only for reference, If you need detailed information, please check here
                    Read more

                    Tantalo (mitologio)

                    Image
                    Ĉi tiu artikolo temas pri mitologio. Por aliaj signifoj vidu la paĝon Tantalo (apartigilo) . Tantalo , laŭ Gioacchino Assereto. Laŭ la helena mitologio, Tantalo (el la greka Τάνταλος , ŝancelanta aŭ tre malfeliĉa ) estis filo de Zeŭso kaj la oceanidino Pluto, laŭ la plej divastigata versio. Li estis reĝo de Frigio [1] aŭ de la monto Sipilo, en Lidio (Malgrandazio) [2] kaj post sia morto li suferis eternan torturadon ĉe la Tartaro, la plej profunda parto de Hadeso, rezervita por la plej perversaj homoj. Tamen, kelkaj rakontoj asertas ke li estis reĝo de Argo aŭ de Korinto [3] , aŭ ke li migris norden el la monto Sipilo por akiri la tronon de Paflagonio, el kie Ilo forpelis lin pro tio, ke Tantalo forkaptis kaj seksperfortis lian fraton Ganimedon [4] , kaj tiel ofendis la diojn. Verŝajne li ankaŭ regis urbon en okcidenta Anatolio, kiun oni nomis Tantalis [5] pro li, aŭ Sipilo , pro la monto sur kies flanko ankoraŭ staras ĝiaj ruinaĵoj [6] . Laŭ la geografiisto Strabo,...
                    Read more

                    Franz Schubert

                    Image
                    Franz Schubert Franz Schubert, pentrita en 1875 fare de Wilhelm August Rieder Bazaj informoj Naskiĝo 31-an de januaro 1797 ( 1797-01-31 ) en Himmelpfortgrund Morto 19-an de novembro 1828 ( 1828-11-19 ) (31-jara) en Vieno Ĝenroj Klasika muziko Profesio komponisto Aktivaj jaroj 1808-1828 v   •   d   •   r Franz SCHUBERT ([ˈfʁants ˈʃuːbɛɐ̯t], esperante Franz' Ŝuberto ; naskiĝis la 31-an de januaro 1797 je Himmelpfortgrund en Vieno, kiu nun estas parto de la 9-a viena distrikto Alsergrund, kaj mortis la 19-an de novembro 1828 en Vieno) estis aŭstra komponisto de klasika kaj fruromantisma muziko. Li komponis ĉirkaŭ 600 lidojn (germanaj kantoj) kaj estas nomata la "reĝo de lido". Enhavo 1 Vivo 1.1 Infaneco 1.2 Junaĝo 1.3 Zelis 1.4 Jaroj de maturiĝo kaj matureco 1.5 La lastaj jaroj 2 Postefiko kaj akceptado 3 Honoradoj 4 Verkoj 4.1 Eklezia muziko 4.2 Dram...
                    Read more