Note that by definition, $f$ and $g$ are bijective functions. Let $f^{-1}:\{a, b, c\} \rightarrow(1,2,3\}$  and $g^{-1}:$ $\{$ apple, ball, cat $\}$ $\rightarrow\{a, b, c\}$ be defined as $f^{-1}\{a\}=1,\, f^{-1}\{b\}=2$, $f^{-1}\{c\}=3$, $g^{-1}$  $\{$ apple $\}=a, \,g^{-1}$ $\{$ ball $\}=b$ and $g^{-1} $ $\{$ cat $\}=c$

It is easy to verify that $f^{-1}$ of $=I_{(1,2,3)}, \,fof^{-1}=I_{\{c, b, c)},\, g^{-1}$ og $=I_{(a, b, c)}$ and $gog^{-1}=I_{D}$ where, $D =\{$ apple, ball, cat  $\} $ . Now, $gof:$  $\{1,2,3\} \rightarrow $ $\{$ apple, ball, cat $\}$ is given by $gof\,(1)=$ apple, $gof(2)=$ ball, $gof(3)=$ cat.

We can define $(gof)^{-1}: \{$ apple, ball , cat $\}$  $\rightarrow\{1,2,3\}$ by $(gof)^{-1}$ (apple) $=1,(gof)^{-1}$ (ball) $=2$ and $(gof)^{-1}$ (cat) $=3$ . It is easy to see that $(gof)^{-1}o(gof)$ $=I_{(1,2,3)}$ and $(gof)o(gof)^{-1}=I_{D} .$

Thus, we have seen that $f,\, g$ and $g$ of are invertible.

Now, $f^{-1}og ^{-1}$ (apple) $=f^{-1}(g^{-1}$ (apple)) $=f^{-1}(a)=1=(gof)^{-1}$ (apple)

$f^{-1}og^{-1} $ (ball) $=f^{-1}(g^{-1}$ (ball)) $=f^{-1}(b)=2=(gof)^{-1}$ (ball) and

$f^{-1} og ^{-1}$ ( cat ) $=f^{-1}(g^{-1}$ ( cat )) $=f^{-1}(c)=3=(gof)^{-1}$ (cat)

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

The above result is true in general situation also.