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

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          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









          $endgroup$













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            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






























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