Let S = \mathbb{R} \ { 1 , -1}. Find a mapping f : S → S such that f\circ f=-\operatorname{id}_{S}. (Hint: try mapping ]—1, 1 [ to its complement in S.)
Consider the mapping f:S\rightarrow S defined by
f(x)=\left\{\begin{array}{c c c}{{1/x}}&{{\mathrm{if}}}&{{x\in{\displaystyle]-1,\,1}[{~~and}\,x\neq0;}}\\ {{0}}&{{\mathrm{if}}}&{{x=0;}}\\ {{-1/x}}&{{\mathrm{if}}}&{{x\not\in{\displaystyle]-1,\,1[}.}}\end{array}\right.It is readily seen that for every x\in S we have f\left[f(x)\right]=-x. The graph of f is shown in Fig. S3.41.