# Show that there exists a neighborhood $U$ of $(0,1)$ such that the restriction $g:U \rightarrow g[U]$ is invertible

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.

• Often, you do not want to construct this neighbourhood explicitly. You are happy that you know it exists... – Mushu Nrek Jul 16 at 22:17

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).
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 $$(0,1)$$.