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How to prove that $f(x)=fracxsqrtx$ is smooth?


Show that an open subset of a smooth manifold is again smooth.Smooth chart in what sense?Proof that this is a smooth manifoldSmooth coverings are open maps proof verificationWhat does it take for a smooth homeomorphism to be a diffeomorphism?Construct a diffeomorphism $psi: B_1 to epsilontext-neighborhood of K$, where $K$ is a subset of a smooth manifold.Diffeomorphism, from class $C^1$ to $C^k$Abstract Smooth Manifolds vs Embedded Smooth ManifoldsSmooth curves that are submanifoldsMaximal Smooth Atlas For Stereographic Projection













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I have a function $f(x)=fracxsqrtx$ on the open ball $B_1(0)$ in $mathbbR^n$. I have to show that this function is a diffeomorphism between $mathbbR^n$ and $B_1(0)$. I have already proved that the function is bijective with inverse $fracxsqrtx$ and it is continuous. How to prove that the function is smooth?



Thanks!










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    1












    $begingroup$


    I have a function $f(x)=fracxsqrtx$ on the open ball $B_1(0)$ in $mathbbR^n$. I have to show that this function is a diffeomorphism between $mathbbR^n$ and $B_1(0)$. I have already proved that the function is bijective with inverse $fracxsqrtx$ and it is continuous. How to prove that the function is smooth?



    Thanks!










    share|cite|improve this question











    $endgroup$














      1












      1








      1





      $begingroup$


      I have a function $f(x)=fracxsqrtx$ on the open ball $B_1(0)$ in $mathbbR^n$. I have to show that this function is a diffeomorphism between $mathbbR^n$ and $B_1(0)$. I have already proved that the function is bijective with inverse $fracxsqrtx$ and it is continuous. How to prove that the function is smooth?



      Thanks!










      share|cite|improve this question











      $endgroup$




      I have a function $f(x)=fracxsqrtx$ on the open ball $B_1(0)$ in $mathbbR^n$. I have to show that this function is a diffeomorphism between $mathbbR^n$ and $B_1(0)$. I have already proved that the function is bijective with inverse $fracxsqrtx$ and it is continuous. How to prove that the function is smooth?



      Thanks!







      differential-geometry smooth-manifolds






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      edited May 7 at 13:33









      Asaf Karagila

      310k33443776




      310k33443776










      asked May 7 at 8:56









      sakasaka

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      504




















          2 Answers
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          $begingroup$

          You can apply the theorem that the sum, product, quotient (whenever defined), square root (of a positive smooth function), and composition of smooth functions are smooth. In your examples, all the component functions $x$, $|x|^2$, $1 pm |x|^2$, and so on are all smooth.






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            2












            $begingroup$

            Just use the fact that compositions of smooth functions is smooth and ratio of two smooth functions is smooth as long as the denominator never vanishes.






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              2 Answers
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              4












              $begingroup$

              You can apply the theorem that the sum, product, quotient (whenever defined), square root (of a positive smooth function), and composition of smooth functions are smooth. In your examples, all the component functions $x$, $|x|^2$, $1 pm |x|^2$, and so on are all smooth.






              share|cite|improve this answer









              $endgroup$

















                4












                $begingroup$

                You can apply the theorem that the sum, product, quotient (whenever defined), square root (of a positive smooth function), and composition of smooth functions are smooth. In your examples, all the component functions $x$, $|x|^2$, $1 pm |x|^2$, and so on are all smooth.






                share|cite|improve this answer









                $endgroup$















                  4












                  4








                  4





                  $begingroup$

                  You can apply the theorem that the sum, product, quotient (whenever defined), square root (of a positive smooth function), and composition of smooth functions are smooth. In your examples, all the component functions $x$, $|x|^2$, $1 pm |x|^2$, and so on are all smooth.






                  share|cite|improve this answer









                  $endgroup$



                  You can apply the theorem that the sum, product, quotient (whenever defined), square root (of a positive smooth function), and composition of smooth functions are smooth. In your examples, all the component functions $x$, $|x|^2$, $1 pm |x|^2$, and so on are all smooth.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered May 7 at 9:06









                  Loring TuLoring Tu

                  1912




                  1912





















                      2












                      $begingroup$

                      Just use the fact that compositions of smooth functions is smooth and ratio of two smooth functions is smooth as long as the denominator never vanishes.






                      share|cite|improve this answer









                      $endgroup$

















                        2












                        $begingroup$

                        Just use the fact that compositions of smooth functions is smooth and ratio of two smooth functions is smooth as long as the denominator never vanishes.






                        share|cite|improve this answer









                        $endgroup$















                          2












                          2








                          2





                          $begingroup$

                          Just use the fact that compositions of smooth functions is smooth and ratio of two smooth functions is smooth as long as the denominator never vanishes.






                          share|cite|improve this answer









                          $endgroup$



                          Just use the fact that compositions of smooth functions is smooth and ratio of two smooth functions is smooth as long as the denominator never vanishes.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered May 7 at 9:04









                          Kavi Rama MurthyKavi Rama Murthy

                          82k53673




                          82k53673



























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