Confusion with implicit function theorem

Question

According to my understanding of the implicit function theorem, if $F(y_0,x_0)=c$ and $\frac{dF}{dy}$ at $(y_0,x_0)$ is nonzero then there is a unique function $y=F(x)$ for $x \in B$ where B is an open ball around $x_0$. The part that is confusing to me is the uniqueness. As the common example of a circle shows ($x^2+y^2=c^2$), for every open ball around $x_0$ there are still two functions $y=\sqrt{x^2-c^2}$ and $y=-\sqrt{x^2-c^2}$. What am I missing here?

Answer 1

What you're missing is the fact that we first fix some $(x_0,y_0)$. In particular, if we choose $(x_0,y_0)$ to be a point on the upper half-circle, then you'll get the positive solution $y=\sqrt{x^2-c^2}$ and if you choose $(x_0,y_0)$ on the lower half-circle you get the negative solution $y=-\sqrt{x^2-c^2}$. The key thing here is that both of them cannot cross $(x_0,y_0)$ at the same time, and so you do in fact get a unique solution as the theorem tells you.

Answer 2

For all the points $(x_0, y_0)$ on the circle and in the strict upper plan, your circle can be approximated by $y = \sqrt{x^2 - c^2}$ and for all the points $(x_0, y_0)$ on the circle and in the strict lower plan, your circle can be approximated by $y = -\sqrt{x^2 - c^2}$.

Your confusion comes from the fact that the development depends on your chosen point. In the first case, all $y$ are negative, and are therefore around a negative $y_0$. Same for the second one.