Does there exist a continous surjective map between $mathbb R^2to S^1$












3












$begingroup$



Does there exist a continuous surjective map between $mathbb R^2to S^1$?




I had find such map with domain $mathbb R^2/${$0$}.



But I do not able to find if we insert 0 there?



Any help will be appreciated






real-analysis general-topology continuity examples-counterexamples







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  • $begingroup$
    Did you mean $mathbf{S}^2$?
    $endgroup$
    – Will M.
    Jan 6 at 4:49
















3












$begingroup$



Does there exist a continuous surjective map between $mathbb R^2to S^1$?




I had find such map with domain $mathbb R^2/${$0$}.



But I do not able to find if we insert 0 there?



Any help will be appreciated






real-analysis general-topology continuity examples-counterexamples







share|cite|improve this question











$endgroup$












  • $begingroup$
    Did you mean $mathbf{S}^2$?
    $endgroup$
    – Will M.
    Jan 6 at 4:49














3












3








3


0



$begingroup$



Does there exist a continuous surjective map between $mathbb R^2to S^1$?




I had find such map with domain $mathbb R^2/${$0$}.



But I do not able to find if we insert 0 there?



Any help will be appreciated






real-analysis general-topology continuity examples-counterexamples







share|cite|improve this question











$endgroup$





Does there exist a continuous surjective map between $mathbb R^2to S^1$?




I had find such map with domain $mathbb R^2/${$0$}.



But I do not able to find if we insert 0 there?



Any help will be appreciated






real-analysis general-topology continuity examples-counterexamples




real-analysis general-topology continuity examples-counterexamples






share|cite|improve this question















share|cite|improve this question













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edited Jan 6 at 8:24









Henno Brandsma

112k348120




112k348120










asked Jan 6 at 4:42









MathLoverMathLover

53710




53710












  • $begingroup$
    Did you mean $mathbf{S}^2$?
    $endgroup$
    – Will M.
    Jan 6 at 4:49


















  • $begingroup$
    Did you mean $mathbf{S}^2$?
    $endgroup$
    – Will M.
    Jan 6 at 4:49
















$begingroup$
Did you mean $mathbf{S}^2$?
$endgroup$
– Will M.
Jan 6 at 4:49




$begingroup$
Did you mean $mathbf{S}^2$?
$endgroup$
– Will M.
Jan 6 at 4:49










3 Answers
3






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6












$begingroup$

How about $f(x,y)=(cos x,sin x)$?






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    2












    $begingroup$

    R^2 to R projection and then from R to circle the exponential map.






    share|cite|improve this answer









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      1












      $begingroup$

      Take $pi:mathbb{R}^2to mathbb{R}$ given by $pi(x,y)=x$, followed by $phi:mathbb{R}to S^1$ given by $phi(t)=e^{2pi i t}$. Both of these maps are continuous and it's easy to see that their composition yields a surjection. So, $phicirc pi:mathbb{R}^2to S^1$ is a continuous surjection.






      share|cite|improve this answer









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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        6












        $begingroup$

        How about $f(x,y)=(cos x,sin x)$?






        share|cite|improve this answer









        $endgroup$


















          6












          $begingroup$

          How about $f(x,y)=(cos x,sin x)$?






          share|cite|improve this answer









          $endgroup$
















            6












            6








            6





            $begingroup$

            How about $f(x,y)=(cos x,sin x)$?






            share|cite|improve this answer









            $endgroup$



            How about $f(x,y)=(cos x,sin x)$?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 6 at 4:45









            SmileyCraftSmileyCraft

            3,571518




            3,571518























                2












                $begingroup$

                R^2 to R projection and then from R to circle the exponential map.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  R^2 to R projection and then from R to circle the exponential map.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    R^2 to R projection and then from R to circle the exponential map.






                    share|cite|improve this answer









                    $endgroup$



                    R^2 to R projection and then from R to circle the exponential map.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 6 at 4:46









                    NeelNeel

                    567314




                    567314























                        1












                        $begingroup$

                        Take $pi:mathbb{R}^2to mathbb{R}$ given by $pi(x,y)=x$, followed by $phi:mathbb{R}to S^1$ given by $phi(t)=e^{2pi i t}$. Both of these maps are continuous and it's easy to see that their composition yields a surjection. So, $phicirc pi:mathbb{R}^2to S^1$ is a continuous surjection.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Take $pi:mathbb{R}^2to mathbb{R}$ given by $pi(x,y)=x$, followed by $phi:mathbb{R}to S^1$ given by $phi(t)=e^{2pi i t}$. Both of these maps are continuous and it's easy to see that their composition yields a surjection. So, $phicirc pi:mathbb{R}^2to S^1$ is a continuous surjection.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Take $pi:mathbb{R}^2to mathbb{R}$ given by $pi(x,y)=x$, followed by $phi:mathbb{R}to S^1$ given by $phi(t)=e^{2pi i t}$. Both of these maps are continuous and it's easy to see that their composition yields a surjection. So, $phicirc pi:mathbb{R}^2to S^1$ is a continuous surjection.






                            share|cite|improve this answer









                            $endgroup$



                            Take $pi:mathbb{R}^2to mathbb{R}$ given by $pi(x,y)=x$, followed by $phi:mathbb{R}to S^1$ given by $phi(t)=e^{2pi i t}$. Both of these maps are continuous and it's easy to see that their composition yields a surjection. So, $phicirc pi:mathbb{R}^2to S^1$ is a continuous surjection.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 6 at 4:45









                            Antonios-Alexandros RobotisAntonios-Alexandros Robotis

                            10.5k41741




                            10.5k41741






























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