Consider $f:\{1,2,3\}\to \{a,b,c\}$ and $g:\{a,b,c\}\to \left\{\text{apple, ball, cat}\right\}$ defined as $f\left(1\right)=a,f\left(2\right)=b,f\left(3\right)=c,g\left(a\right)=apple,g\left(b\right)=ball$ and $g\left(c\right)=cat$. Show that $f.g$ and $gof$ are invertible. Find out $f^{-1},g^{-1}$ and $(gof{)}^{-1}$ and show that $(gof{)}^{-1}=f^{-1}0g^{-1}$.

Checking one-one, onto and finding $f^{-1}$ for $f$.
$f:\{1,2,3\}\to \{a,b,c\}$
$f\left(1\right)=a,f\left(2\right)=b,f\left(3\right)=c$
Since,all elements have distinct image, $f$ is one-one.
Now, $f=\left\{\right(1,a),(2,b),(3,c\left)\right\}$
Since, every image has a pre-image, $f$ is onto.
Since $f$is one-one and onto, $f$ is invertible
So, $f^{-1}=\left\{\right(a,1),(b,2),(c,3\left)\right\}$

Checking one-one, onto and finding $g^{-1}$ for $g$.
$g:\{a,b,c\}\to \left\{\text{apple, ball, cat}\right\}$
$g\left(a\right)=apple,g\left(b\right)=ball,g\left(c\right)=cat,$
Since, all elements have distinct image, $g$ is one-one.
Since $g$ is one-one and onto, $g$ is invertible
$g=\left\{\text{a, apple, b, ball, c, cat}\right\}$
$\therefore g^{-1}=\left\{\text{(apple,a) (ball, b) (cat, c)}\right\}$

Checking one-one, onto and finding for $(gof{)}^{-1}$.
$f:\{1,2,3\}\to \{a,b,c\},g:\{a,b,c\}\to \left\{\text{apple, ball, cat}\right\}$
So, $gof$ will be,
$gof=\left\{\right(1,apple),(2,ball),(3,cat\left)\right\}$
Since, all elements have distinct image, $gof$ is one-one.
Since, every image has a pre-image.
$gof$ is onto.
$gof$ is one-one and onto.
$\therefore gof$ is invertible.
$gof=\left\{\right(1,apple),(2,ball),(3,cat\left)\right\}$
$\therefore (gof{)}^{-1}=\left\{\right(apple,1),(ball,2),(cat,3\left)\right\}$

Finding $f^{-1}0g^{-1}$.
$f^{-1}=\left\{\right(a,1),(b,2),(c,3\left)\right\}\&$
$g^{-1}=\left\{\right(apple,a),(ball,b),(cat,c\left)\right\}$
$\therefore f^{-1}0g^{-1}=\left\{\right(apple,1),(ball,2),(cat,3\left)\right\}$
We found
$(gof{)}^{-1}=\left\{\right(apple,1),(ball,2),(cat,3\left)\right\}$
$\therefore (gof{)}^{-1}=f^{-1}0g^{-1}$