Are the manifolds $N=(mathbb{R},text{Id})$ and $M=(mathbb{R},xmapsto x^{frac{1}{3}})$ diffeomorphic?












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Are the manifolds $N=(mathbb{R},text{Id})$ and $M=(mathbb{R},xmapsto x^{frac{1}{3}})$ diffeomorphic?




I have already shown that $text{Id}: N rightarrow M$ is a homeomorphism but not a diffeomorphism.










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    Are the manifolds $N=(mathbb{R},text{Id})$ and $M=(mathbb{R},xmapsto x^{frac{1}{3}})$ diffeomorphic?




    I have already shown that $text{Id}: N rightarrow M$ is a homeomorphism but not a diffeomorphism.










    share|cite|improve this question









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      $begingroup$



      Are the manifolds $N=(mathbb{R},text{Id})$ and $M=(mathbb{R},xmapsto x^{frac{1}{3}})$ diffeomorphic?




      I have already shown that $text{Id}: N rightarrow M$ is a homeomorphism but not a diffeomorphism.










      share|cite|improve this question









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      Are the manifolds $N=(mathbb{R},text{Id})$ and $M=(mathbb{R},xmapsto x^{frac{1}{3}})$ diffeomorphic?




      I have already shown that $text{Id}: N rightarrow M$ is a homeomorphism but not a diffeomorphism.







      differential-geometry smooth-manifolds diffeomorphism






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      asked Dec 20 '18 at 12:14









      Jens WagemakerJens Wagemaker

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          Yes. Try to show that $F:Nto M$ given by $F(x)=x^3$ is a diffeomorphism.






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            Yes, they are diffeomorphic. Consider the map$$begin{array}{ccc}N&longrightarrow&M\x&mapsto&x^3.end{array}$$It's a diffeomorphism.






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              $begingroup$

              Yes. Try to show that $F:Nto M$ given by $F(x)=x^3$ is a diffeomorphism.






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                $begingroup$

                Yes. Try to show that $F:Nto M$ given by $F(x)=x^3$ is a diffeomorphism.






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                  $begingroup$

                  Yes. Try to show that $F:Nto M$ given by $F(x)=x^3$ is a diffeomorphism.






                  share|cite|improve this answer









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                  Yes. Try to show that $F:Nto M$ given by $F(x)=x^3$ is a diffeomorphism.







                  share|cite|improve this answer












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                  share|cite|improve this answer










                  answered Dec 20 '18 at 12:18









                  positrón0802positrón0802

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                      Yes, they are diffeomorphic. Consider the map$$begin{array}{ccc}N&longrightarrow&M\x&mapsto&x^3.end{array}$$It's a diffeomorphism.






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                        $begingroup$

                        Yes, they are diffeomorphic. Consider the map$$begin{array}{ccc}N&longrightarrow&M\x&mapsto&x^3.end{array}$$It's a diffeomorphism.






                        share|cite|improve this answer









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                          $begingroup$

                          Yes, they are diffeomorphic. Consider the map$$begin{array}{ccc}N&longrightarrow&M\x&mapsto&x^3.end{array}$$It's a diffeomorphism.






                          share|cite|improve this answer









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                          Yes, they are diffeomorphic. Consider the map$$begin{array}{ccc}N&longrightarrow&M\x&mapsto&x^3.end{array}$$It's a diffeomorphism.







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                          answered Dec 20 '18 at 12:17









                          José Carlos SantosJosé Carlos Santos

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