Solution:

$f=\{(1, a),(2, b),(3, c)\}$ and $g=\{(a$, apple $),(b$, ball $),(c$, cat $)\}$

Clearly, $f$ and $g$ are bijections.

So, $f$ and $g$ are invertible.

Now,

$f^{-1}=\{(a, 1),(b, 2),(c, 3)\}$ and $g^{-1}=\{($ apple,$a)$, (ball, $b$ ), (cat, $\left.c)\right\}$

So, $f^{-1} o g^{-1}=\{($ apple, 1$),($ ball, 2$),($ cat, 3$)\} \quad \ldots(1)$

$f:\{1,2,3\} \rightarrow\{a, b, c\}$ and $g:\{a, b, c\} \rightarrow\{$ apple, ball, cat $\}$

So, $g o f:\{1,2,3\} \rightarrow\{$ apple, ball, cat $\}$

$\Rightarrow(g \circ f)(1)=g(f(1))=g(a)=$ apple

$(g o f)(2)=g(f(2))=g(b)=$ ball

and $(g o f)(3)=g(f(3))=g(c)=$ cat

$\therefore g o f=\{(1$, apple $),(2$, ball $),(3$, cat $)\}$        

Clearly, gofis a bijection.

So, gof is invertible.

$(\text { gof })^{-1}=\{($ apple, 1$),($ ball, 2$),($ cat, 3$)\}$             ...(2)

From $(1)$ and $(2)$, we get:

$(g o f)^{-1}=f^{-1} o g^{-1}$