Piecewise smooth function , Jacobian and locally invertible?

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Say $f$ is a piecewise smooth function that is $f(x,y;chi)=f_1(x,y;chi)$ if $yleq g(x;chi)$
and
$f(x,y;chi)=f_2(x,y;chi)$ if $y geq g(x;chi)$



where $g$ is a $C^2$ function and $chi$ a parameter.



Now I am trying to understand this statement -



If the product of $det(Df_1)_(0,1;0)$ and $det(Df_2)_(0,1;0)$ is positive, then $f$ is locally invertible



Well if the product of the determinant is positive then neither of the determinants is non-zero and hence locally invertible by Inverse function theorem



But I am thinking whether saying this would be wrong?



If the product of $det(Df_1)_(0,1;0)$ and $det(Df_2)_(0,1;0)$ is negative or non-zero, then $f$ is locally invertible?



Also Can we generalize this?




calculus real-analysis jacobian piecewise-continuity inverse-function-theorem




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    up vote
    1
    down vote

    favorite
    1












    Say $f$ is a piecewise smooth function that is $f(x,y;chi)=f_1(x,y;chi)$ if $yleq g(x;chi)$
    and
    $f(x,y;chi)=f_2(x,y;chi)$ if $y geq g(x;chi)$



    where $g$ is a $C^2$ function and $chi$ a parameter.



    Now I am trying to understand this statement -



    If the product of $det(Df_1)_(0,1;0)$ and $det(Df_2)_(0,1;0)$ is positive, then $f$ is locally invertible



    Well if the product of the determinant is positive then neither of the determinants is non-zero and hence locally invertible by Inverse function theorem



    But I am thinking whether saying this would be wrong?



    If the product of $det(Df_1)_(0,1;0)$ and $det(Df_2)_(0,1;0)$ is negative or non-zero, then $f$ is locally invertible?



    Also Can we generalize this?




    calculus real-analysis jacobian piecewise-continuity inverse-function-theorem




    share|cite|improve this question
























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      Say $f$ is a piecewise smooth function that is $f(x,y;chi)=f_1(x,y;chi)$ if $yleq g(x;chi)$
      and
      $f(x,y;chi)=f_2(x,y;chi)$ if $y geq g(x;chi)$



      where $g$ is a $C^2$ function and $chi$ a parameter.



      Now I am trying to understand this statement -



      If the product of $det(Df_1)_(0,1;0)$ and $det(Df_2)_(0,1;0)$ is positive, then $f$ is locally invertible



      Well if the product of the determinant is positive then neither of the determinants is non-zero and hence locally invertible by Inverse function theorem



      But I am thinking whether saying this would be wrong?



      If the product of $det(Df_1)_(0,1;0)$ and $det(Df_2)_(0,1;0)$ is negative or non-zero, then $f$ is locally invertible?



      Also Can we generalize this?




      calculus real-analysis jacobian piecewise-continuity inverse-function-theorem




      share|cite|improve this question














      Say $f$ is a piecewise smooth function that is $f(x,y;chi)=f_1(x,y;chi)$ if $yleq g(x;chi)$
      and
      $f(x,y;chi)=f_2(x,y;chi)$ if $y geq g(x;chi)$



      where $g$ is a $C^2$ function and $chi$ a parameter.



      Now I am trying to understand this statement -



      If the product of $det(Df_1)_(0,1;0)$ and $det(Df_2)_(0,1;0)$ is positive, then $f$ is locally invertible



      Well if the product of the determinant is positive then neither of the determinants is non-zero and hence locally invertible by Inverse function theorem



      But I am thinking whether saying this would be wrong?



      If the product of $det(Df_1)_(0,1;0)$ and $det(Df_2)_(0,1;0)$ is negative or non-zero, then $f$ is locally invertible?



      Also Can we generalize this?





      calculus real-analysis jacobian piecewise-continuity inverse-function-theorem





      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 28 at 7:49









      Martin Sleziak

      43.6k6113260




      43.6k6113260










      asked Aug 25 at 10:26









      BAYMAX

      2,54521021




      2,54521021

























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