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

(N/A) The function $f: \{1, 2, 3\} \rightarrow \{a, b, c\}$ is defined as $f(1) = a, f(2) = b, f(3) = c$.
To find $f^{-1}$,we define a function $g: \{a, b, c\} \rightarrow \{1, 2, 3\}$ such that $g(a) = 1, g(b) = 2, g(c) = 3$.
We check the compositions:
$(f \circ g)(a) = f(g(a)) = f(1) = a$
$(f \circ g)(b) = f(g(b)) = f(2) = b$
$(f \circ g)(c) = f(g(c)) = f(3) = c$
Thus,$f \circ g = I_Y$,where $Y = \{a, b, c\}$.
$(g \circ f)(1) = g(f(1)) = g(a) = 1$
$(g \circ f)(2) = g(f(2)) = g(b) = 2$
$(g \circ f)(3) = g(f(3)) = g(c) = 3$
Thus,$g \circ f = I_X$,where $X = \{1, 2, 3\}$.
Since $f \circ g = I_Y$ and $g \circ f = I_X$,the inverse of $f$ exists and $f^{-1} = g$.
Therefore,$f^{-1}: \{a, b, c\} \rightarrow \{1, 2, 3\}$ is defined by $f^{-1}(a) = 1, f^{-1}(b) = 2, f^{-1}(c) = 3$.
Now,to find $(f^{-1})^{-1}$,we find the inverse of $g$. Let $h: \{1, 2, 3\} \rightarrow \{a, b, c\}$ be defined as $h(1) = a, h(2) = b, h(3) = c$.
We check the compositions:
$(g \circ h)(1) = g(h(1)) = g(a) = 1$
$(g \circ h)(2) = g(h(2)) = g(b) = 2$
$(g \circ h)(3) = g(h(3)) = g(c) = 3$
Thus,$g \circ h = I_X$.
$(h \circ g)(a) = h(g(a)) = h(1) = a$
$(h \circ g)(b) = h(g(b)) = h(2) = b$
$(h \circ g)(c) = h(g(c)) = h(3) = c$
Thus,$h \circ g = I_Y$.
Since $g \circ h = I_X$ and $h \circ g = I_Y$,the inverse of $g$ exists and $g^{-1} = h$. Since $g = f^{-1}$,we have $(f^{-1})^{-1} = h$. Since $h = f$,it follows that $(f^{-1})^{-1} = f$.