The inverse of a parametrization is differentiable.

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Hi. I'm thinking about a statement in the example above (do Carmo):



If $mathbfx:UsubsetmathbbR^2to S$ is a parametrization, $mathbfx^-1:mathbfx(U)tomathbbR^2$ is differentiable.



========



Being a parametrization, $mathbfx$ is certainly differentiable and has an inverse $mathbfx^-1$. Is this strong enough to guarantee that $mathbfx^-1$ is differentiable? How could this happen? Thanks.




differential-geometry




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  • Do you know the inverse function theorem?
    – Osama Ghani
    Jul 15 at 8:19










  • Yes. Maybe I should identify the definition of differentiability of $mathbfx^-1$.
    – iamokay
    Jul 16 at 15:05















up vote
1
down vote

favorite












enter image description here



Hi. I'm thinking about a statement in the example above (do Carmo):



If $mathbfx:UsubsetmathbbR^2to S$ is a parametrization, $mathbfx^-1:mathbfx(U)tomathbbR^2$ is differentiable.



========



Being a parametrization, $mathbfx$ is certainly differentiable and has an inverse $mathbfx^-1$. Is this strong enough to guarantee that $mathbfx^-1$ is differentiable? How could this happen? Thanks.




differential-geometry




share|cite|improve this question



















  • Do you know the inverse function theorem?
    – Osama Ghani
    Jul 15 at 8:19










  • Yes. Maybe I should identify the definition of differentiability of $mathbfx^-1$.
    – iamokay
    Jul 16 at 15:05













up vote
1
down vote

favorite









up vote
1
down vote

favorite











enter image description here



Hi. I'm thinking about a statement in the example above (do Carmo):



If $mathbfx:UsubsetmathbbR^2to S$ is a parametrization, $mathbfx^-1:mathbfx(U)tomathbbR^2$ is differentiable.



========



Being a parametrization, $mathbfx$ is certainly differentiable and has an inverse $mathbfx^-1$. Is this strong enough to guarantee that $mathbfx^-1$ is differentiable? How could this happen? Thanks.




differential-geometry




share|cite|improve this question











enter image description here



Hi. I'm thinking about a statement in the example above (do Carmo):



If $mathbfx:UsubsetmathbbR^2to S$ is a parametrization, $mathbfx^-1:mathbfx(U)tomathbbR^2$ is differentiable.



========



Being a parametrization, $mathbfx$ is certainly differentiable and has an inverse $mathbfx^-1$. Is this strong enough to guarantee that $mathbfx^-1$ is differentiable? How could this happen? Thanks.





differential-geometry





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 15 at 2:22









iamokay

537




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  • Do you know the inverse function theorem?
    – Osama Ghani
    Jul 15 at 8:19










  • Yes. Maybe I should identify the definition of differentiability of $mathbfx^-1$.
    – iamokay
    Jul 16 at 15:05

















  • Do you know the inverse function theorem?
    – Osama Ghani
    Jul 15 at 8:19










  • Yes. Maybe I should identify the definition of differentiability of $mathbfx^-1$.
    – iamokay
    Jul 16 at 15:05
















Do you know the inverse function theorem?
– Osama Ghani
Jul 15 at 8:19




Do you know the inverse function theorem?
– Osama Ghani
Jul 15 at 8:19












Yes. Maybe I should identify the definition of differentiability of $mathbfx^-1$.
– iamokay
Jul 16 at 15:05





Yes. Maybe I should identify the definition of differentiability of $mathbfx^-1$.
– iamokay
Jul 16 at 15:05
















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