# Can we get a contradiction of the inverse function theorem?

If $f:D\subset\mathbb{R}^2\rightarrow\mathbb{R}^2$ is defined by the rule $$f(x,y)=(x^2+y^2,x^2-y^2)$$ where $D$ is the square $$D=\{(x,y)\in\mathbb{R}^2|0\leq x\leq1,0\leq y\leq1\}$$ Then we see that $f$ is one-to-one but is not onto and hence $f$ does not have an inverse.

However, we find the Jacobian of $f$ is $-8xy$ and from the inverse function theorem, we know that wherever this is not zero, then we have an inverse.

Does not this contradict the inverse function theorem?

This is the inverse function theorem from page 253 of Tromba and Marsden vector calculus 5th edition.

Inverse Function Theorem Let $U\subset\mathbb{R}^n$ be open and let $f_1:U\to\mathbb{R},\ldots,f_n:U\to\mathbb{R}$ have continuous partial derivatives. Consider the equations: \begin{align*} f_1(x_1,\ldots,x_n)&=y_1\\ \vdots\\ f_n(x_1,\ldots,x_n)&=y_n \end{align*} near a given solution $\mathbf{x}_0$, $\mathbf{y}_0$. If $J(f)(\mathbf{x}_0)$ is nonzero, then the above equation can be solved uniquely as $\mathbf{x}=g(\mathbf{y})$ for $\mathbf{x}$ near $\mathbf{x}_0$ and $\mathbf{y}$ near $\mathbf{y}_0$. Moreover, the function $g$ has continuous partial derivatives.

• I haven't looked at your specific function, but your complaint seems to be that it can't have an inverse because it's not onto. But that's not what the inverse function theorem says. The inverse function theorem says there is an inverse function in a neighborhood of any point in the image, not the range. – Callus - Reinstate Monica Mar 29 '16 at 11:58
• Do you know the statement of the Inverse function theorem correctly? – Paladin Mar 29 '16 at 12:00
• @Neel: I have added the statement of inverse function theorem in my question. It may be that I do not understand it properly. – user326822 Mar 29 '16 at 12:11
• @Callus: Do you want to make your comment in the form of an answer? – user326822 Mar 29 '16 at 12:13
• @user326822 Note in the statement it's mentioned that solution exists near $(x_0,y_0)$ .....so you might not get a global solution instead you can always solve it locally in a unique way whenever the conditions are satisfied.... – Paladin Mar 29 '16 at 12:56

Maybe it is easier, just for understanding, to reformulate the inverse function theorem in terms of regular points of the domain of a given differentiable real function. A regular point is a point of the domain where the differential of the map is non singular, practically speaking is that point where the Jacobian matrix is non singular. The inverse function theorem works for regular points and states that you can choose a neighborhood of the regular point and the restriction of the given differentiable function at that neighborhood is a local diffeomorphism. So in these terms in your example there are no contradiction of the inverse function theorem in fact your function fails to be a local diffeomorphism only at the origin of the Plane, but it is a diffeomorphism when you restrict your domain at this set $\{\ (x,y) \in \mathbb{R^2} : 0<x \leq 1 , 0<y\leq1 \}$