Prove that $T: mathbb{R}^2 rightarrow mathbb{R}^2$ is a bijection












4












$begingroup$


I have to prove that the following is a bijection: $ T: mathbb{R}^2 rightarrow mathbb{R}^2 $, with $T(x,y)= left(
begin{array}{c}
5x + sin(y)\
5y + arctan(x)
end{array}
right)$



Now, to prove it is surjective I composed it with $x = tan(z)$ obtaining:



$ T^1colon left]-pi /2, pi /2 right[times mathbb{R} rightarrow mathbb{R}^2 $, with $T^1(z,y)= left(
begin{array}{c}
5tan(z) + sin(y)\
5y + z
end{array}
right)
$



Then, if we consider a generic $(z_0,y_0)$ we have $5y+z=y_0 Rightarrow z = y_0 - 5y$ and $5 tan(y_0 - 5y) + sin(y)=g(y)$. Considering the limits for $g(y)$ at the inf and sup of its domain are $+ infty$ and $-infty$ and the continuity of $g(y)$ we prove the existence of $Y$ such that $g(Y) = z_0$ and then $Z = y_0 - 5Y$ and we've found a couple $(Y,Z)$ such that $T^1(Y,Z)=(z_0,y_0)$.



As for the injectivity I've noticed that, using the fact that sin and arctan are limited functions, we get $T(x,y)=T(x',y') Rightarrow |x-x'|< frac{2}{5}$ and $|y-y'|< frac{2 pi}{5}$ but I can't do any better.



My questions are:



1) Is there an easier or better proof of surjectivity? (given that mine is correct...if not please point out where I did wrong)



2) Can somebody give me hints to prove the injectivity?



Thank you very much in advance!






functions







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    I have to prove that the following is a bijection: $ T: mathbb{R}^2 rightarrow mathbb{R}^2 $, with $T(x,y)= left(
    begin{array}{c}
    5x + sin(y)\
    5y + arctan(x)
    end{array}
    right)$



    Now, to prove it is surjective I composed it with $x = tan(z)$ obtaining:



    $ T^1colon left]-pi /2, pi /2 right[times mathbb{R} rightarrow mathbb{R}^2 $, with $T^1(z,y)= left(
    begin{array}{c}
    5tan(z) + sin(y)\
    5y + z
    end{array}
    right)
    $



    Then, if we consider a generic $(z_0,y_0)$ we have $5y+z=y_0 Rightarrow z = y_0 - 5y$ and $5 tan(y_0 - 5y) + sin(y)=g(y)$. Considering the limits for $g(y)$ at the inf and sup of its domain are $+ infty$ and $-infty$ and the continuity of $g(y)$ we prove the existence of $Y$ such that $g(Y) = z_0$ and then $Z = y_0 - 5Y$ and we've found a couple $(Y,Z)$ such that $T^1(Y,Z)=(z_0,y_0)$.



    As for the injectivity I've noticed that, using the fact that sin and arctan are limited functions, we get $T(x,y)=T(x',y') Rightarrow |x-x'|< frac{2}{5}$ and $|y-y'|< frac{2 pi}{5}$ but I can't do any better.



    My questions are:



    1) Is there an easier or better proof of surjectivity? (given that mine is correct...if not please point out where I did wrong)



    2) Can somebody give me hints to prove the injectivity?



    Thank you very much in advance!






    functions







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      1



      $begingroup$


      I have to prove that the following is a bijection: $ T: mathbb{R}^2 rightarrow mathbb{R}^2 $, with $T(x,y)= left(
      begin{array}{c}
      5x + sin(y)\
      5y + arctan(x)
      end{array}
      right)$



      Now, to prove it is surjective I composed it with $x = tan(z)$ obtaining:



      $ T^1colon left]-pi /2, pi /2 right[times mathbb{R} rightarrow mathbb{R}^2 $, with $T^1(z,y)= left(
      begin{array}{c}
      5tan(z) + sin(y)\
      5y + z
      end{array}
      right)
      $



      Then, if we consider a generic $(z_0,y_0)$ we have $5y+z=y_0 Rightarrow z = y_0 - 5y$ and $5 tan(y_0 - 5y) + sin(y)=g(y)$. Considering the limits for $g(y)$ at the inf and sup of its domain are $+ infty$ and $-infty$ and the continuity of $g(y)$ we prove the existence of $Y$ such that $g(Y) = z_0$ and then $Z = y_0 - 5Y$ and we've found a couple $(Y,Z)$ such that $T^1(Y,Z)=(z_0,y_0)$.



      As for the injectivity I've noticed that, using the fact that sin and arctan are limited functions, we get $T(x,y)=T(x',y') Rightarrow |x-x'|< frac{2}{5}$ and $|y-y'|< frac{2 pi}{5}$ but I can't do any better.



      My questions are:



      1) Is there an easier or better proof of surjectivity? (given that mine is correct...if not please point out where I did wrong)



      2) Can somebody give me hints to prove the injectivity?



      Thank you very much in advance!






      functions







      share|cite|improve this question











      $endgroup$




      I have to prove that the following is a bijection: $ T: mathbb{R}^2 rightarrow mathbb{R}^2 $, with $T(x,y)= left(
      begin{array}{c}
      5x + sin(y)\
      5y + arctan(x)
      end{array}
      right)$



      Now, to prove it is surjective I composed it with $x = tan(z)$ obtaining:



      $ T^1colon left]-pi /2, pi /2 right[times mathbb{R} rightarrow mathbb{R}^2 $, with $T^1(z,y)= left(
      begin{array}{c}
      5tan(z) + sin(y)\
      5y + z
      end{array}
      right)
      $



      Then, if we consider a generic $(z_0,y_0)$ we have $5y+z=y_0 Rightarrow z = y_0 - 5y$ and $5 tan(y_0 - 5y) + sin(y)=g(y)$. Considering the limits for $g(y)$ at the inf and sup of its domain are $+ infty$ and $-infty$ and the continuity of $g(y)$ we prove the existence of $Y$ such that $g(Y) = z_0$ and then $Z = y_0 - 5Y$ and we've found a couple $(Y,Z)$ such that $T^1(Y,Z)=(z_0,y_0)$.



      As for the injectivity I've noticed that, using the fact that sin and arctan are limited functions, we get $T(x,y)=T(x',y') Rightarrow |x-x'|< frac{2}{5}$ and $|y-y'|< frac{2 pi}{5}$ but I can't do any better.



      My questions are:



      1) Is there an easier or better proof of surjectivity? (given that mine is correct...if not please point out where I did wrong)



      2) Can somebody give me hints to prove the injectivity?



      Thank you very much in advance!






      functions




      functions






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      edited Dec 20 '18 at 16:37









      Arthur Jr.

      476




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      asked Aug 14 '13 at 11:10









      GivAlzGivAlz

      30317




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          2 Answers
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          $begingroup$

          One way to prove that $T , : , mathbb{R}^{2} , rightarrow , mathbb{R}^{2}$ is bijective could be to use the global inverse theorem (also called Hadamard-Levy theorem, see http://arxiv.org/pdf/1305.5930v2.pdf - theorem 1.1) because $mathrm{det} , mathrm{Jac}(T,(x,y))$ never vanishes and $Vert T(x,y) Vert rightarrow +infty$ as $Vert (x,y) Vert rightarrow +infty$.



          However, there may be a better answer to the question.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Check that you satisfy the conditions given by the Theorem in the pdf of Jibounet.



            For the Jacobian part, compute it at $(x, y)$



            $$
            J = begin{bmatrix}5 & cos(y) \ frac{1}{1+x^2} & 5 end{bmatrix},
            $$



            and check if the determinant of $J$ is zero for some pair $(x, y)$.



            $$
            text{det}(J) = 25 - frac{cos(y)}{1+x^2} = frac{25x^2 + 25 - cos(y)}{1+x^2}neq 0, , forall x, y in mathbb{R},
            $$
            since $|cos(y)|leq 1$.






            share|cite|improve this answer











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              2 Answers
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              active

              oldest

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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              One way to prove that $T , : , mathbb{R}^{2} , rightarrow , mathbb{R}^{2}$ is bijective could be to use the global inverse theorem (also called Hadamard-Levy theorem, see http://arxiv.org/pdf/1305.5930v2.pdf - theorem 1.1) because $mathrm{det} , mathrm{Jac}(T,(x,y))$ never vanishes and $Vert T(x,y) Vert rightarrow +infty$ as $Vert (x,y) Vert rightarrow +infty$.



              However, there may be a better answer to the question.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                One way to prove that $T , : , mathbb{R}^{2} , rightarrow , mathbb{R}^{2}$ is bijective could be to use the global inverse theorem (also called Hadamard-Levy theorem, see http://arxiv.org/pdf/1305.5930v2.pdf - theorem 1.1) because $mathrm{det} , mathrm{Jac}(T,(x,y))$ never vanishes and $Vert T(x,y) Vert rightarrow +infty$ as $Vert (x,y) Vert rightarrow +infty$.



                However, there may be a better answer to the question.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  One way to prove that $T , : , mathbb{R}^{2} , rightarrow , mathbb{R}^{2}$ is bijective could be to use the global inverse theorem (also called Hadamard-Levy theorem, see http://arxiv.org/pdf/1305.5930v2.pdf - theorem 1.1) because $mathrm{det} , mathrm{Jac}(T,(x,y))$ never vanishes and $Vert T(x,y) Vert rightarrow +infty$ as $Vert (x,y) Vert rightarrow +infty$.



                  However, there may be a better answer to the question.






                  share|cite|improve this answer









                  $endgroup$



                  One way to prove that $T , : , mathbb{R}^{2} , rightarrow , mathbb{R}^{2}$ is bijective could be to use the global inverse theorem (also called Hadamard-Levy theorem, see http://arxiv.org/pdf/1305.5930v2.pdf - theorem 1.1) because $mathrm{det} , mathrm{Jac}(T,(x,y))$ never vanishes and $Vert T(x,y) Vert rightarrow +infty$ as $Vert (x,y) Vert rightarrow +infty$.



                  However, there may be a better answer to the question.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 14 '13 at 12:05









                  jibounetjibounet

                  5,6981232




                  5,6981232























                      1












                      $begingroup$

                      Check that you satisfy the conditions given by the Theorem in the pdf of Jibounet.



                      For the Jacobian part, compute it at $(x, y)$



                      $$
                      J = begin{bmatrix}5 & cos(y) \ frac{1}{1+x^2} & 5 end{bmatrix},
                      $$



                      and check if the determinant of $J$ is zero for some pair $(x, y)$.



                      $$
                      text{det}(J) = 25 - frac{cos(y)}{1+x^2} = frac{25x^2 + 25 - cos(y)}{1+x^2}neq 0, , forall x, y in mathbb{R},
                      $$
                      since $|cos(y)|leq 1$.






                      share|cite|improve this answer











                      $endgroup$


















                        1












                        $begingroup$

                        Check that you satisfy the conditions given by the Theorem in the pdf of Jibounet.



                        For the Jacobian part, compute it at $(x, y)$



                        $$
                        J = begin{bmatrix}5 & cos(y) \ frac{1}{1+x^2} & 5 end{bmatrix},
                        $$



                        and check if the determinant of $J$ is zero for some pair $(x, y)$.



                        $$
                        text{det}(J) = 25 - frac{cos(y)}{1+x^2} = frac{25x^2 + 25 - cos(y)}{1+x^2}neq 0, , forall x, y in mathbb{R},
                        $$
                        since $|cos(y)|leq 1$.






                        share|cite|improve this answer











                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Check that you satisfy the conditions given by the Theorem in the pdf of Jibounet.



                          For the Jacobian part, compute it at $(x, y)$



                          $$
                          J = begin{bmatrix}5 & cos(y) \ frac{1}{1+x^2} & 5 end{bmatrix},
                          $$



                          and check if the determinant of $J$ is zero for some pair $(x, y)$.



                          $$
                          text{det}(J) = 25 - frac{cos(y)}{1+x^2} = frac{25x^2 + 25 - cos(y)}{1+x^2}neq 0, , forall x, y in mathbb{R},
                          $$
                          since $|cos(y)|leq 1$.






                          share|cite|improve this answer











                          $endgroup$



                          Check that you satisfy the conditions given by the Theorem in the pdf of Jibounet.



                          For the Jacobian part, compute it at $(x, y)$



                          $$
                          J = begin{bmatrix}5 & cos(y) \ frac{1}{1+x^2} & 5 end{bmatrix},
                          $$



                          and check if the determinant of $J$ is zero for some pair $(x, y)$.



                          $$
                          text{det}(J) = 25 - frac{cos(y)}{1+x^2} = frac{25x^2 + 25 - cos(y)}{1+x^2}neq 0, , forall x, y in mathbb{R},
                          $$
                          since $|cos(y)|leq 1$.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Aug 14 '13 at 12:18

























                          answered Aug 14 '13 at 12:11







                          user51196





































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