If functions $f:A \to B$ and $g:B \to A$
satisfy $gof = {I_A}$, then show that $f$ is
one-one and $g$ is onto.

It is given that,, $f:A \to B$ and $g:B \to A$ satisfy

$gof$ $= {I_A}$

$gof$ $= {I_A}$

$\Rightarrow$ $gof\left\{ {f\left( {{x_1}} \right)} \right\} = gof\left\{ {f\left( {{x_2}} \right)} \right\}$

$\Rightarrow$ $g\left( {{x_1}} \right) = g\left( {{x_2}} \right)$.

[$gof$ $= {I_A}$]

$\therefore$ ${x_1} = {x_2}$

Hence we can say that, $f$ is one-one and $g$ is onto.