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It is given that
$f : \left \{ 1,2,3 \right \}\rightarrow \left \{ a,b,c \right \}$

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

Now,, lets define a function g :
$\left \{ a,b,c \right \}\rightarrow \left \{ 1,2,3 \right \}$ such that

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

Now,

$(fog)(a) = f(g(a)) = f(1) = a$
Similarly,

$(fog)(b) = f(g(b)) = f(2) = b$

$(fog)(c) = f(g(c)) = f(3) = c$

And

$(gof)(1) = g(f(1)) = g(a) = 1$

$(gof)(2) = g(f(2)) = g(b) = 2$

$(gof)(3) = g(f(3)) = g(c) = 3$

Hence, $gof = I_X$ and $fog = I_Y$ , where $X = \left \{ 1,2,3 \right \}$ and $Y = \left \{ a,b,c \right \}$

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

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

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

Now, we need to find the inverse of $f^{-1}$ ,

Therefore, lets define $h: \left \{ 1,2,3 \right \}\rightarrow \left \{ a,b,c \right \}$ such that

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

Now,

$(goh)(1) = g(h(1)) = g(a) = 1$

$(goh)(2) = g(h(2)) = g(b) = 2$

$(goh)(3) = g(h(3)) = g(c) = 3$

Similarly,

$(hog)(a) = h(g(a)) = h(1) = a$

$(hog)(b) = h(g(b)) = h(2) = b$

$(hog)(c) = h(g(c)) = h(3) = c$

Hence, $goh = I_X$ and $hog = I_Y$ , where $X = \left \{ 1,2,3 \right \}$ and $Y = \left \{ a,b,c \right \}$

Therefore, inverse of $g^{-1} = (f^{-1})^{-1}$ exists and $g^{-1} = (f^{-1})^{-1} = h$

$\Rightarrow h = f$

Therefore, $(f^{-1})^{-1} = f$

Hence proved

Posted by

Gautam harsolia

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