Homeomorphism between region above parabola and $ mathbb R ^2$












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Define the set $ X = { (x,y) in mathbb R ^2 : x^2 < y }. $
Construct a homeomorphism between $ X $ and $ mathbb R ^2 $.




Graphically, $ X $ is the region above the parabola $y = x^2$, not including the boundary lines. Since it's contained entirely in the upper half plane, I thought of trying something involving logarithms and considered $Phi: X to mathbb R ^2, ; (x,y) to (x, ln sqrt y)$. As it stands, $f$ is injective and also continuous as the composition of continuous maps on the domain of definition but I wasn't able to show surjectivity. For any $ (x,y) in mathbb R ^2 $ we have $(x,y) = Phi (x, e^{2y} ) $, but the latter isn't necessarily contained in $X$ so I'm not sure how to proceed (or even if this kind of function is the right form to be considering).






general-topology differential-geometry







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    0












    $begingroup$



    Define the set $ X = { (x,y) in mathbb R ^2 : x^2 < y }. $
    Construct a homeomorphism between $ X $ and $ mathbb R ^2 $.




    Graphically, $ X $ is the region above the parabola $y = x^2$, not including the boundary lines. Since it's contained entirely in the upper half plane, I thought of trying something involving logarithms and considered $Phi: X to mathbb R ^2, ; (x,y) to (x, ln sqrt y)$. As it stands, $f$ is injective and also continuous as the composition of continuous maps on the domain of definition but I wasn't able to show surjectivity. For any $ (x,y) in mathbb R ^2 $ we have $(x,y) = Phi (x, e^{2y} ) $, but the latter isn't necessarily contained in $X$ so I'm not sure how to proceed (or even if this kind of function is the right form to be considering).






    general-topology differential-geometry







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      Define the set $ X = { (x,y) in mathbb R ^2 : x^2 < y }. $
      Construct a homeomorphism between $ X $ and $ mathbb R ^2 $.




      Graphically, $ X $ is the region above the parabola $y = x^2$, not including the boundary lines. Since it's contained entirely in the upper half plane, I thought of trying something involving logarithms and considered $Phi: X to mathbb R ^2, ; (x,y) to (x, ln sqrt y)$. As it stands, $f$ is injective and also continuous as the composition of continuous maps on the domain of definition but I wasn't able to show surjectivity. For any $ (x,y) in mathbb R ^2 $ we have $(x,y) = Phi (x, e^{2y} ) $, but the latter isn't necessarily contained in $X$ so I'm not sure how to proceed (or even if this kind of function is the right form to be considering).






      general-topology differential-geometry







      share|cite|improve this question











      $endgroup$





      Define the set $ X = { (x,y) in mathbb R ^2 : x^2 < y }. $
      Construct a homeomorphism between $ X $ and $ mathbb R ^2 $.




      Graphically, $ X $ is the region above the parabola $y = x^2$, not including the boundary lines. Since it's contained entirely in the upper half plane, I thought of trying something involving logarithms and considered $Phi: X to mathbb R ^2, ; (x,y) to (x, ln sqrt y)$. As it stands, $f$ is injective and also continuous as the composition of continuous maps on the domain of definition but I wasn't able to show surjectivity. For any $ (x,y) in mathbb R ^2 $ we have $(x,y) = Phi (x, e^{2y} ) $, but the latter isn't necessarily contained in $X$ so I'm not sure how to proceed (or even if this kind of function is the right form to be considering).






      general-topology differential-geometry




      general-topology differential-geometry






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited Jan 10 at 22:58







      backstrapp

















      asked Jan 10 at 22:46









      backstrappbackstrapp

      12819




      12819






















          2 Answers
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          You can do it in two steps:




          1. Consider the map$$begin{array}{rccc}psicolon&X&longrightarrow&mathbb{R}^2\&(x,y)&mapsto&(x,y-x^2).end{array}$$Then $psi$ is a homeomorphism from $X$ onto $mathbb{R}times(0,infty)$.

          2. Now, define a homeomorphism from $mathbb{R}times(0,infty)$ onto $mathbb{R}^2$ and compose both homeomorphisms.






          share|cite|improve this answer











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          • $begingroup$
            Very slick approach!
            $endgroup$
            – backstrapp
            Jan 10 at 23:03



















          0












          $begingroup$

          Hint: For the strategy $(x,y)mapsto(x,star)$ to work, you'll want a function of both $x$ and $y$ in the $star$.






          share|cite|improve this answer









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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            6












            $begingroup$

            You can do it in two steps:




            1. Consider the map$$begin{array}{rccc}psicolon&X&longrightarrow&mathbb{R}^2\&(x,y)&mapsto&(x,y-x^2).end{array}$$Then $psi$ is a homeomorphism from $X$ onto $mathbb{R}times(0,infty)$.

            2. Now, define a homeomorphism from $mathbb{R}times(0,infty)$ onto $mathbb{R}^2$ and compose both homeomorphisms.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Very slick approach!
              $endgroup$
              – backstrapp
              Jan 10 at 23:03
















            6












            $begingroup$

            You can do it in two steps:




            1. Consider the map$$begin{array}{rccc}psicolon&X&longrightarrow&mathbb{R}^2\&(x,y)&mapsto&(x,y-x^2).end{array}$$Then $psi$ is a homeomorphism from $X$ onto $mathbb{R}times(0,infty)$.

            2. Now, define a homeomorphism from $mathbb{R}times(0,infty)$ onto $mathbb{R}^2$ and compose both homeomorphisms.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Very slick approach!
              $endgroup$
              – backstrapp
              Jan 10 at 23:03














            6












            6








            6





            $begingroup$

            You can do it in two steps:




            1. Consider the map$$begin{array}{rccc}psicolon&X&longrightarrow&mathbb{R}^2\&(x,y)&mapsto&(x,y-x^2).end{array}$$Then $psi$ is a homeomorphism from $X$ onto $mathbb{R}times(0,infty)$.

            2. Now, define a homeomorphism from $mathbb{R}times(0,infty)$ onto $mathbb{R}^2$ and compose both homeomorphisms.






            share|cite|improve this answer











            $endgroup$



            You can do it in two steps:




            1. Consider the map$$begin{array}{rccc}psicolon&X&longrightarrow&mathbb{R}^2\&(x,y)&mapsto&(x,y-x^2).end{array}$$Then $psi$ is a homeomorphism from $X$ onto $mathbb{R}times(0,infty)$.

            2. Now, define a homeomorphism from $mathbb{R}times(0,infty)$ onto $mathbb{R}^2$ and compose both homeomorphisms.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 11 at 16:43

























            answered Jan 10 at 22:54









            José Carlos SantosJosé Carlos Santos

            165k22132235




            165k22132235












            • $begingroup$
              Very slick approach!
              $endgroup$
              – backstrapp
              Jan 10 at 23:03


















            • $begingroup$
              Very slick approach!
              $endgroup$
              – backstrapp
              Jan 10 at 23:03
















            $begingroup$
            Very slick approach!
            $endgroup$
            – backstrapp
            Jan 10 at 23:03




            $begingroup$
            Very slick approach!
            $endgroup$
            – backstrapp
            Jan 10 at 23:03











            0












            $begingroup$

            Hint: For the strategy $(x,y)mapsto(x,star)$ to work, you'll want a function of both $x$ and $y$ in the $star$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Hint: For the strategy $(x,y)mapsto(x,star)$ to work, you'll want a function of both $x$ and $y$ in the $star$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Hint: For the strategy $(x,y)mapsto(x,star)$ to work, you'll want a function of both $x$ and $y$ in the $star$.






                share|cite|improve this answer









                $endgroup$



                Hint: For the strategy $(x,y)mapsto(x,star)$ to work, you'll want a function of both $x$ and $y$ in the $star$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 10 at 22:54









                Chris CulterChris Culter

                21.4k43887




                21.4k43887






























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