Idempotent solutions to the implict function theorem other than the identity?

Question

I am interested in the following problem. Assume that an (anti)symmetric function $g:\mathbb{R}^2 \to \mathbb{R}$ satisfies the implicit function theorem. That is, $g(x,y) = \pm g(y,x)$ and $g(x,y)=0$ means that we can find $h: \mathbb{R} \to \mathbb{R}$ such that $x=h(y)$. I want to know whether there are functions $g$ that admit idempotent solutions, i.e. $x=h(y)$ with $h^2(y) = h(y)$. As an example, take $g(x,y) = x - y$. It is clear that we have $x = h(y)= y$ and thus $h$ is idempotent. Is anyone able to construct other examples where $h$ is not the identity and perhaps provide necessary/sufficient on $g$ for this to be possible? Thanks.

Answer 1

Examples are abundant. If $x_0$ is a point such that $g(x_0,x_0)=0$ and $g_y(x_0,x_0)\neq 0$ then $g_x(x_0,x_0)\neq 0$ the Implicit Function Theorem, there is a unique function $y=\phi(x), \phi(x_0)=x_0$, and the symmetry implies that $\phi$ coincides with its inverse. For example, if $x\in (0,1),y\in(0,1)$ then: $$x^2+y^2-1=0, \quad y(x)=\sqrt{1-x^2},\quad y(y(x))=x.$$