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Let $g: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be defined by $g(x,y)=(2ye^{2x},xe^y)$. Show that there exists a neighborhood $U$ of $(0,1)$ such that the restriction $g:U \rightarrow g[U]$ is invertible, and $g^{-1} \in C^{\infty}(g[U];\mathbb{R}^2)$.

Here is the theorem I am thinking about (Inverse Function Theorem): Let $W \subseteq \mathbb{R}^n$ be open, and let $f \in \mathbb{C}^r(W; \mathbb{R}^n), r \le 1$. If $a \in W$ is a point such that $Df(a)$ is an invertible matrix, then there exist open sets $U \subseteq W$ and $V \subseteq \mathbb{R}^n$ such that $a \in U$ and the restriction $f:U \rightarrow V$ is invertible with $f^{-1} \in \mathbb{C}^r(V; \mathbb{R}^m)$.

The related lemma is: let $U \subseteq \mathbb{R}^n$ be open, and let $f \in \mathbb{C}^1(U;\mathbb{R}^n)$, and let $a \in U$. If $Df(a)$ is an invertible matrix, then there exists $\alpha, \epsilon>0$ such that $\lVert f(x_0)-f(x_1)\rVert \leq \sigma \lVert x_0-x_1 \rVert$ for all $x_0,x_1 \in \mathbb{C}(\alpha, \epsilon)$.

I am still stuck in constructing the neighborhood for this question. Any help is appreciated.

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    $\begingroup$ Often, you do not want to construct this neighbourhood explicitly. You are happy that you know it exists... $\endgroup$ – Mushu Nrek Jul 16 at 22:17
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Hint: Compute $Df$, and evaluate it at the point $a=(0,1)$. Is $Df(a)$ an invertible matrix? If yes, then the inverse function theorem tells us that such a neighborhood exists, and you're done! No need to actually construct such a neighborhood, it is sufficient to know that it's there (at least, that's what your questions seems to ask).

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The Jacobian of $g$ at $(0,1)$ is $\left[\begin{smallmatrix}4&2\\e&0\end{smallmatrix}\right]$, which is invertible, since its determinant is different from $0$. So, the Inverse Function Theorem tells you that $g$ is invertible in some neighborhood of $(0,1)$.

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