Question

(a) Let $\mathit{f}\mathbf{}\mathbf{:}\mathbf{}\mathit{B}\mathbf{\to }\mathit{C}$androle="math" localid="1659170400956" $\mathit{g}\mathbf{}\mathbf{:}\mathbf{}\mathit{C}\mathbf{\to }\mathit{D}$be functions such that$\mathit{g}\mathbf{}\mathit{o}\mathbf{}\mathit{f}$is surjective. Prove that is surjective.

(b) Give an example of the situation in part (a) in which f is not surjective.

Short answer

Thus,

(a) it is proved that g is surjective.

(b)$f\left(x\right)=x-4$ is surjective.

Step-by-step solution

(a) Showing that g is injective

It is given that g of is surjective.

Let $z\in D$.

Then sinceg of is surjective, there exists $x\in B$such that,

$$\begin{gathered} \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right) \\ =z \end{gathered}$$

Thus, let$y=f\left(x\right)\in C$then,

$g\left(y\right)=z$

Therefore, g is surjective.

Hence, it is proved that g is surjective.

(b) Example of the situation in part (a) in which f is not surjective.

It is given that g (x) = tan x for $x\ne kx+\frac{\pi }{2}$, and $g\left(kx+\frac{\pi }{2}\right)=0,k=1,2,3,....$.

Now, when $f\left(x\right)={\tan }^{-1}x$ so f maps !onto $\left(-\frac{\mathrm{\pi }}{2},\frac{\pi }{2}\right)$, f is not surjective but g of is surjective.

Hence,f is not surjective but g of is surjective.