Involution as solution of Babbage's equation

Example

Babbage's functional equation asks for maps $f \colon X \to X$ with $f(f(x)) = x$ for all $x \in X$. Its solutions are exactly the involutions of $X$, and also exactly the iterative roots of order $2$ of the identity.

Take $X = \R \setminus \set{-1}$ and $f(x) = \frac{1 - x}{1 + x}$. First, $f$ maps $X$ into $X$: the equation $\frac{1 - x}{1 + x} = -1$ would give $1 - x = -1 - x$, which is impossible. Second, for $x \in X$, $$f(f(x)) = \frac{1 - \frac{1-x}{1+x}}{1 + \frac{1-x}{1+x}} = \frac{(1 + x) - (1 - x)}{(1 + x) + (1 - x)} = \frac{2x}{2} = x.$$ So $f$ is an involution and solves Babbage's equation. Other solutions on suitable sets are $x \mapsto c - x$, $x \mapsto 1/x$ and the identity.