Question 3.44: For mappings f , g : R → R and every λ ∈ R define the mappin......

(a) Take, for example,  $f=\mathrm{id}_{\mathbb{R}}\ \mathrm{and}\ g=-\mathrm{id}_{\mathbb{R}}.$  Then (f + g)(x) = 0 for every  $x\in\mathbb{R}$  and so f + g is not a bijection.

Consider now  $f=\operatorname{id}_{\mathbb{R}}$  and define  $g:\mathbb{R}\to\mathbb{R}$  by

Then g is a bijection; its graph is shown in Fig. S3.42.

Now

so f·g is not a bijection. For this f and g we also have

The equation  $x+1/x=1$  has no solution in  $\mathbb{R}_{*}$  so there is  no  $x\in\mathbb{R}$  such that  $({{f}}+g)(x)=1\,.$  Thus f + g is not a bijection.

(b) Suppose that f is a bijection. Then for  $\lambda\neq0$   we have

and so  $\lambda f$  is injective. Also, since f is surjective, given any  $y\in\mathbb{R},$  there exists  $t\in\mathbb{R}$  with  $f(t)=y/\lambda.$  Then  $(\lambda f)(t)=y$  and hence   $\lambda f$  is also surjective. This shows that if f is a bijection then so is   $\lambda f$  for every  $\lambda\neq0.$   Suppose now that   $\lambda f$  is a bijection with  $\lambda\neq0.$  Applying the above result to   $\lambda f$   and  $\mu=1/\lambda,$   it follows that  $\mu(\lambda f)=(1/\lambda)(\lambda f)=f$   is also a bijection.

(c) Consider, for example, the mappings  $f,g:\mathbb{R}\to\mathbb{R}{\mathrm{~given~by}}f(x)=x^{2},$  $g(x)=x+1.$  We have

so  $\textstyle[f g]$  is a bijection.

(d) We have

and similarly

Adding these together, we obtain the required result.