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I've proven that the function $f: U=(0,\infty)\times \mathbb R \rightarrow \mathbb R^2$ given by $f(x,y) = (x, y^3 + xy)$ is injective and surjective ($f(U) = U$), so it is bijective.

I've computed $\frac {\partial f_1}{\partial x} = 1$, $\frac {\partial f_1}{\partial y} = 0$, $\frac {\partial f_2}{\partial x} = y$ and $\frac {\partial f_2}{\partial y} = 3y^2+x$.

Can someone tell me how to apply The Inverse Function Theorem to prove $f$ is a diffeomorphism ? ($f$ is smooth is easy to see, I've shown it is bijective). I need only to verify it has a smooth inverse function $f^{-1}: U \rightarrow U$.

I find the theorem confusing to apply.

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You compute the determinant of the Jacobian matrix (of which you computed already its components). The result is $1\times(2y^2+x)-0\times y=2y^2+x$. Then you can see that this is never zero on $U$ ($x>0$ and $y^2\geq0$).

The inverse function theorem gives you the differentiability of the inverse.

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  • $\begingroup$ So this function $g$ one gets from TIFT must be equal to $f^{-1}$ by the property that $f(g(v))=v$ for all $v$ in the small open region in $f(U)$ ? And since $g$ is smooth, this implies $f^{-1}$ is smooth ? $\endgroup$ – Shuzheng Feb 21 '15 at 18:10
  • $\begingroup$ Yes, if $f$ has an inverse and $f(g(v))=v$ then $g$ is that inverse. $\endgroup$ – user218170 Feb 21 '15 at 18:13
  • $\begingroup$ Ahh, and since the region is open around this function $g$, it satisfies the definition of taking the derivative :) Thanks ! $\endgroup$ – Shuzheng Feb 21 '15 at 18:21

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