Question

Let $f:A\to B$ and $g:B\to C$ be the bijective functions. Then $(gof{)}^{-1}$ is

• A. $fog$
• B. $f^{-1}o{g}^{-1}$
• C. $gof$
• D. $g^{-1}o{f}^{-1}$

Answer 1

The correct option is B $f^{-1}o{g}^{-1}$

Let $f:A\to B$ be one-one onto and $g:B\to C$ be one-one onto.



We first show that $gof$ is one-one onto.

$\left(gof\right)$ is one-one, since

$\left(gof\right)\left(x_1\right)=\left(gof\right)\left(x_2\right)$

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

$\Rightarrow f\left(x_1\right)=f\left(x_2\right)$ $[\because g\text{is one-one}]$

$\Rightarrow x_1=x_2$ $[\because f\text{is one-one}]$

Let $z\in C.$ Then, $g$ being onto, these exists

$y\in B$ such that $g\left(y\right)=z.$

Now, $f$ being onto, these exists $x\in A$ such that $f\left(x\right)=y.$

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

$=g\left\{f\right(x\left)\right\}$ $[\because y=f(x\left)\right]$

$=\left(gof\right)\left(x\right)$

Thus, for each $z\in C,$ there exists $x\in A$ such that
$\left(gof\right)\left(x\right)=z.$

$\therefore \left(gof\right)$ is onto.

Thus $\left(gof\right)$ is one-one onto.

Now, $f\left(x\right)=y\Rightarrow f^{-1}\left(y\right)=x.$

And, $g\left(y\right)=z\Rightarrow g^{-1}\left(z\right)=y.$

Also, $\left(gof\right)\left(x\right)=z\Rightarrow \left(gof{)}^{-1}\right(z)=x.$

$\therefore \left(f^{-1}o{g}^{-1}\right)\left(z\right)=f^{-1}\left\{g^{-1}\right(z\left)\right\}$

$=f^{-1}\left(y\right)$ $[\because g^{-1}(z)=y]$

$=x$ $[\because f^{-1}(y)=x]$

$=\left(gof{)}^{-1}\right(z).$

Hence, $(gof{)}^{-1}=(f^{-1}o{g}^{-1}).$

Therefore, option (a) is correct.