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Let $A$ and $B$ be sets. Show that $\mathrm{f}: \mathrm{A} \times \mathbf{B} \rightarrow \mathbf{B} \times \mathbf{A}$ such that $\mathbf{f}(\mathbf{a}, \mathbf{b})=(\mathrm{b}, \mathbf{a})$ is bijective function.
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We have $f:(A \times B) \rightarrow B \times A$ such that $f(a, b)=b, a$
(a) $\quad \therefore \mathrm{f}\left(\mathrm{a}_1, \mathrm{~b}_1\right)=\left(\mathrm{b}_1, \mathrm{a}_1\right) \quad \mathrm{f}\left(\mathrm{a}_2, \mathrm{~b}_2\right)=\left(\mathrm{b}_2, \mathrm{a}_2\right)$
$$
\mathrm{f}\left(\mathrm{a}_1, \mathrm{~b}_1\right)=\mathrm{f}\left(\mathrm{a}_2, \mathrm{~b}_2\right) \Rightarrow\left(\mathrm{b}_1, \mathrm{a}_1\right)=\left(\mathrm{b}_2, \mathrm{a}_2\right)
$$
$\Rightarrow \mathrm{b}_1=\mathrm{b}_2$ and $\mathrm{a}_1=\mathrm{a}_2 \quad \therefore \mathrm{f}$ is one-one
(b) Every member $(\mathrm{p}, \mathrm{q})$ belonging to its codomain has its pre-image in its domain as $(q, p) f$ is onto. Thus, $f$ is one-one and onto i.e. it is bijective.
(a) $\quad \therefore \mathrm{f}\left(\mathrm{a}_1, \mathrm{~b}_1\right)=\left(\mathrm{b}_1, \mathrm{a}_1\right) \quad \mathrm{f}\left(\mathrm{a}_2, \mathrm{~b}_2\right)=\left(\mathrm{b}_2, \mathrm{a}_2\right)$
$$
\mathrm{f}\left(\mathrm{a}_1, \mathrm{~b}_1\right)=\mathrm{f}\left(\mathrm{a}_2, \mathrm{~b}_2\right) \Rightarrow\left(\mathrm{b}_1, \mathrm{a}_1\right)=\left(\mathrm{b}_2, \mathrm{a}_2\right)
$$
$\Rightarrow \mathrm{b}_1=\mathrm{b}_2$ and $\mathrm{a}_1=\mathrm{a}_2 \quad \therefore \mathrm{f}$ is one-one
(b) Every member $(\mathrm{p}, \mathrm{q})$ belonging to its codomain has its pre-image in its domain as $(q, p) f$ is onto. Thus, $f$ is one-one and onto i.e. it is bijective.
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