Question

Let B and C be nonempty sets. Prove that the function$\mathbf{f}\mathbf{:}\mathbf{B}\mathbf{\times }\mathbf{C}\mathbf{\to }\mathbf{C}\mathbf{\times }\mathbf{B}$ given by$\mathbf{f}\left(x,y\right)\mathbf{=}\left(y,x\right)$ is a bijection.

Short answer

It can be concluded that the exhibit function f is a bijection.

Step-by-step solution

Consider the given statement

It is given that the function$f:B\times C\to C\times B$ is a bijection function. As it is also provided that B and C be nonempty sets and$\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\left(\mathrm{y},\mathrm{x}\right)$

prove the function f is injection

To prove a function f is bijection; then it must be one to one and onto both. For f is bijection,

It is assumed that $\mathrm{f}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=\mathrm{f}\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$then,

$\left({\mathrm{y}}_1,{\mathrm{x}}_1\right)=\mathrm{f}\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$

Therefore ${\mathrm{y}}_1={\mathrm{y}}_2$and ${\mathrm{x}}_1={\mathrm{x}}_2$. Hence, $\mathrm{f}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=\mathrm{f}\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$. So the function f is bijection.

Prove the function f is bijection

It is assumed that there is an arbitary element $\left(\mathrm{y},\mathrm{x}\right)\in \mathrm{C}\times \mathrm{B}$. Evidently

$\left(\mathrm{y},\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)$

Therefore, $\left(\mathrm{y},\mathrm{x}\right)\in \mathrm{Image}\left(\mathrm{f}\right)$,hence f is surjective. So, the provided function is bijection.