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

Toolbox:
• A function $g$ is called inverse of $f:x \to y$, then exists $g:y \to x$ such that $gof=I_x\;and\; fog=I_y$, where $I_x, I_y$ are identify functions.
• To find $(f^{-1})^{-1})$, define $g = f^{-1}$ and find $g^{-1} = h$ and show that $h$ is nothing but $f$
Given a function $f:\{1,2,3\} \to \{a,b,c\}$ given by $f(1)=a\;f(2)=b\;f(3)=c$
$\textbf{Step 1: Calculating}\;\; \mathbf{f^{-1} = g}$
We define $g;\{a,b,c\} \to \{1,2,3\}$ as $g(a)=1;\;g(b)=2\;g(c)=3$
$\Rightarrow$ $(fog)(a)=f(g(a))=f(1)=a$
$\Rightarrow$ $(fog)(b)=f(g(b))=f(2)=b$
$\Rightarrow$ $(fog)(c)=f(g(c))=f(3)=c$
$\Rightarrow$ $(gof)(1)=g(f(1))=g(a)=1$
$\Rightarrow$ $(gof)(2)=g(f(2))=g(b)=2$
$\Rightarrow$ $(gof)(3)=g(f(3))=g(c)=3$
Therefore $gof=I_x\;and\;fog=I_y \Rightarrow$ $f^{-1}=g$
$\Rightarrow f^{-1}:{a,b,c} \to {1,2,3}$ given by $f^{-1}(a), \;f^{-1}(b)=2, \; f^{-1}(c)=3$
$\textbf{Step 2: Calculating}\;\; \mathbf{g^{-1} = h}$
Define $h:\{1,2,3\} \to \{a,b,c\}$ as $h(1)=a\;h(2)=h\;f(3)=c$
$\Rightarrow$ $(goh)(1)=g(h(1))=g(a)=1$
$\Rightarrow$ $(goh)(2)=g(h(2))=g(b)=2$
$\Rightarrow$ $(goh)(3)=g(h(3))=g(c)=3$
$\Rightarrow$ $(hog)(a)=h(g(a))=h(1)=a$
$\Rightarrow$ $(hog)(b)=h(g(b))=h(2)=b$
$\Rightarrow$ $(hog)(c)=h(g(c))=h(3)=b$
Therefore, $yoh=I_x\;and\;hog=I_y \Rightarrow g^{-1} = h$
$\textbf{Step 3: Calculating}\;\; \mathbf{(f^{-1})^{-1}}$
From Step 2 above, $g^{-1}=h$ and from Step 1 above, $f^{-1}=g$.
Therefore $(f^{-1})^{-1} = g^{-1} = h \rightarrow$ $f$ and $h$ are the same function.
Since $h=f$, $\;(f^{-1})^{-1}=f$
edited Mar 19, 2013