Consider $f:\{1,2,3\} \rightarrow\{a, b, c\}$ and $g:\{a, b, c\} \rightarrow\{$ apple, ball, cat $\}$ defined as $f(1)=a, f(2)=b, f(3)=c, g(a)=$ apple, $g(b)=$ ball and $g(c)=$ cat. Show that $f, g$ and $g o f$ are
invertible. Find $f^{-1}, g^{-1}$ and $g \circ f^{-1}$ and show that $(g \circ f)^{-1}=f^{-1} \circ g^{-1}$.
$f=\{(1, a),(2, b),(3, c)\}$ 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:
$(g o f)^{-1}=f^{-1} o g^{-1}$
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