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# Consider f : {1, 2, 3} → {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.

11 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  $(f^{-1})^{-1} = f$.

Views

$f : \{1, 2, 3\} \rightarrow \{a, b, c\}$

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

Let there be a function g such that $g:\left \{ a,b,c \right \} = \left \{ 1,2,3 \right \}$

i.e.  $g (a) = 1$    ,  $g (b) = 2$      and      $g (c) = 3$

Now , we have

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

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

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

And,

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

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

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

$\therefore$        $gof = I_x \, \, \, and \, \, \, fog=I_y$, here   $X =\left \{ 1,2,3 \right \}\, and \, Y=\left \{ a,b,c \right \}$

Hence, f exists and $f^{-1}$ is g.

$f^{-1}:\left \{ a,b,c \right \} = \left \{ 1,2,3 \right \}$

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

Let inverse of $f^{-1}$ be h such that  $h: \{1, 2, 3\} \rightarrow \{a, b, c\}$

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

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

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

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

And

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

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

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

$\therefore$      $goh=I_x \, \, \, and\, \, \, hog=I_y$,  here  $X =\left \{ 1,2,3 \right \}\, and \, Y=\left \{ a,b,c \right \}$

Thus, $g^{-1}= h =( f^{-1})^{-1}$

It is noted that h=f.

Hence,$( f^{-1})^{-1}=h=f$.

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