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I am attempting to solve this problem:

Show that the system of equations $$ \begin{align} x &= u^4 - u + uv + v^2, \\ y &= \cos u + \sin v \end{align} $$ can be solved for $(u,v)$ as a smooth function $F$ of $(x,y)$, in some neighborhood of $(0,0)$, in such a way that $(u,v) = (0,0)$ when $(x,y) = (0,1)$. What is the differential of the resulting function $F$ at $(0,1)$?

If I define $G(u,v) = (u^4 - u + uv + v^2,\cos u + \sin v)$, then $G^{-1} = F$. But $$ \det dG^{-1}(0,0) = \begin{vmatrix} 0 & 0\\ 0 & 1\\ \end{vmatrix} = 0. $$

This means that I cannot use the Inverse Function Theorem to show $G^{-1}$ is smooth near $(0,0)$. I suspect that due to the periodic nature of $y$, there may be a different inverse function that does allow me to use the Inverse Function Theorem. However, the problem gives me $F(0,1) = (0,0)$ so I feel like I should be using $(u,v)=(0,0)$ to find the inverse function.

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    $\begingroup$ You made a mistake while calculating the Jacobian of the function $G$: \begin{equation} d G = \begin{pmatrix} 4u^3-1+v & u + 2v \\ -\sin u & \cos v \end{pmatrix}. \end{equation} Evaluating this at $(u,v) = (0,0)$ we get \begin{equation} dG(0,0) = \begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix} \end{equation} so that $det\,dG^{-1} (0,0) = -1 \neq 0$ and the Inverse Function Theorem applies. $\endgroup$ – pg_star Mar 5 at 3:06
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As pg_star pointed out in the comments, I had calculated the Jacobian of $G$ incorrectly. $$ dG(0,0) = \begin{pmatrix} -1 & 0\\ 0 & 1\\ \end{pmatrix} $$

Because $G(0,0) = (0,1)$, the Inverse Function Theorem says that $dF(0,1)$ is the inverse of $dG(0,0)$, which is $$ \begin{pmatrix} -1 & 0\\ 0 & 1\\ \end{pmatrix} .$$

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