Choose open sets to be balls in implicit function theorem

Question

The conclusion of the implicit function theorem for $\mathbb{R}^n$ says that if $F(a, b) = 0$, then there is a function $g$ from an open $V \subseteq \mathbb{R}^m$ to an open $U \subseteq \mathbb{R}^k$ such that $g(b) = a$, and $F(u, v) = 0$ for $u \in U,\ v \in V$ if and only if $g(v) = u$. I've seen one or two source claim that both $U$ and $V$ can be chosen to be open balls. Is this true? Why? Obviously one or the other can be restricted, but how can both be chosen as open balls simultaneously?

Answer 1

In general they can't.

The following would be a simple example except that we can take the open sets to be $\mathbb{R}^2$, however it illustrates the issue: $F(x,y) = (x_1-y_1,2x_2-y_2)$.

To fix this, take $F(x,y) = (x_1-y_1, x_2-y_1^2)$, and use the $\|\cdot\|_2$ norm for $x$ and $y$. It is clear that $\frac{\partial f(x,y)}{\partial x} = I $ for all $x,y$, and we must have $g(y) = (y_1, y_2^2)$.

If we take $\hat{y} = (\hat{y}_1 , \hat{y}_2)$ and $V=B(\hat{y}, \epsilon)$, then it is clear that $g(V)$ is not of the form $B(\hat{x}, \eta)$ for some $\hat{x},\eta>0$.

Answer 2

The definition of open set and open ball are equivalent since $\mathbb R^n$ is a metric space. Then every open set contains an open ball and every open ball is an open set.

Answer 3

Both sets can be open balls inside of the respective open sets $U,V$ ; they can both be open balls because the restriction of the map to $U,V$ is a diffeomorphism, by the inverse function theorem, since the Jacobian matrix $Jf$ is invertible.