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Question 3.41: Let S =  R \ { 1 , -1}. Find a mapping f : S → S such that f......

Let S =  R\mathbb{R} \ { 1 , -1}. Find a mapping f : S → S such that   f∘f=−id⁡S.f\circ f=-\operatorname{id}_{S}.   (Hint: try mapping ]—1, 1 [ to its complement in S.)

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Consider the mapping  f:S→Sf:S\rightarrow S  defined by

f(x)={1/xifx∈]−1, 1[  and x≠0;0ifx=0;−1/xifx∉]−1, 1[.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∈Sx\in S  we have  f[f(x)]=−x.f\left[f(x)\right]=-x.  The graph of f is shown in Fig. S3.41.

Screenshot 2023-10-16 164249 (1)

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