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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.

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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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