Question

If f : A → A, g : A → A are two bijections, then prove that
(i) fog is an injection
(ii) fog is a surjection

Answer 1

Given: A → A, g : A → A are two bijections.
Then, fog : A → A

(i) Injectivity of fog:
Let x and y be two elements of the domain (A), such that
$$\begin{gathered} \left(fog\right)\left(x\right)=\left(fog\right)\left(y\right) \\ \Rightarrow f\left(g\left(x\right)\right)=f\left(g\left(y\right)\right) \\ \Rightarrow g\left(x\right)=g\left(y\right)\left(\text{As,}\mathit{\text{f}}\text{is one-one}\right) \\ \Rightarrow x=y\left(\text{As,}\mathit{\text{g}}\text{is one-one}\right) \end{gathered}$$
So, fog is an injection.

(ii) Surjectivity of fog:
Let z be an element in the co-domain of fog (A).
$$\begin{gathered} \text{Now,}\mathit{\text{z∈A}}\left(\text{co-domain of}\mathit{\text{f}}\right)\text{and}\mathit{\text{f}}\text{is a surjection.} \\ \text{So,}z\mathit{=}f\left(y\right)\mathit{,}\mathit{}\text{where}\mathit{\text{y∈A}}\left(\mathrm{domain}\mathrm{of}\mathit{}f\right)\mathit{.}\mathit{.}\mathit{.}\left(1\right) \\ \text{Now}\mathit{\text{, y∈A}}\left(\text{co-domain of}\mathit{\text{g}}\right)\text{and}\mathit{\text{g}}\text{is a surjection.} \\ \text{So}\mathit{\text{,}}y\mathit{=}g\left(x\right)\mathit{,}\text{where}\mathit{\text{x∈A}}\left(\mathrm{domain}\mathrm{of}\mathit{}g\right)\mathit{.}\mathit{.}\mathit{.}\left(2\right) \\ \mathrm{From}\left(1\right)\mathrm{and}\left(2\right), \\ z\mathit{=}f\left(y\right)\mathit{=}f\left(g\left(x\right)\right)\mathit{=}\left(fog\right)\left(x\right)\mathit{,}\mathit{}\mathrm{where}x\mathit{\in }A\left(\mathrm{domain}\mathrm{of}fog\right) \end{gathered}$$
So, fog is a surjection.