# Consider f: {1, 2, 3} → {a, b, c} given by f(1) = a, f(2) = b and f(3) = c.

Question:

Consider $t:\{1,2,3\} \rightarrow\{a, b, c\}$ given by $t(1)=a, t(2)=b$ and $t(3)=c$. Find $f^{-1}$ and show that $\left(f^{-1}\right)^{-1}=f$.

Solution:

Function $f:\{1,2,3\} \rightarrow\{a, b, c\}$ is given by,

$f(1)=a, f(2)=b$, and $f(3)=c$

If we define $g:\{a, b, c\} \rightarrow\{1,2,3\}$ as $g(a)=1, g(b)=2, g(c)=3$, then we have:

Thus, the inverse of $f$ exists and $f^{-1}=g$.

$\therefore f^{-1}:\{a, b, c\} \rightarrow\{1,2,3\}$ is given by,

$f^{-1}(a)=1, f^{-1}(b)=2, f^{-1}(c)=3$

Let us now find the inverse of $f^{-1}$ i.e., find the inverse of $g$.

If we define $h:\{1,2,3\} \rightarrow\{a, b, c\}$ as

$h(1)=a, h(2)=b, h(3)=c$, then we have:

Thus, the inverse of $g$ exists and $g^{-1}=h \Rightarrow\left(f^{-1}\right)^{-1}=h$.

It can be noted that $h=f$.

Hence, $\left(f^{-1}\right)^{-1}=f$.