# Consider f : {1, 2, 3} → {a, b, c} given by f (1) = a, f (2) = b, f (3) = c. Find (f ^{- 1})^{- 1} = f

**Solution:**

A function is a process or a relation that associates each element 'a' of a non-empty set A, to a single element 'b' of another non-empty set B.

According to the given problem,

Function f : {1, 2, 3} → {a, b, c} given by

f (1) = a, f (2) = b, f (3) = c.

If we define g : {a, b, c} → {1, 2, 3} as

g (a) = 1, g (b) = 2, g (c) = 3

(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} = {(1, 2, 3)} and Y = {a, b, c}

Thus, the inverse of f exists and f ^{- 1} = g.

⇒ f ^{- }^{1} : {a, b, c} → {1, 2, 3} is given by, f ^{- }^{1} (a) = 1, f ^{- }^{1} (b) = 2, f ^{- }^{1} (c) = 3

We need to find the inverse of f ^{-}^{1} i.e., inverse of g.

If we define h : {1, 2, 3} → {a, b, c} as h (1) = a, h (2) = b, 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

⇒ goh = I_{X} and hog = I_{Y} where X = {(1, 2, 3)} and Y = {a, b, c}

Thus, the inverse of g exists and g^{- }^{1} = h ⇒ (f ^{- }^{1})^{- }^{1} = h.

It can be noted that h = f.

Hence, (f ^{- }^{1})^{- }^{1} = f

NCERT Solutions for Class 12 Maths - Chapter 1 Exercise 1.3 Question 11

## Consider f : {1, 2, 3} → {a, b, c} given by f (1) = a, f (2) = b, f (3) = c. Find (f ^{- 1})^{- 1} = f.

**Summary:**

For the function f : {1, 2, 3} → {a, b, c} given by f (1) = a, f (2) = b, f (3) = c. It can be concluded that the inverse of g exists and g^{- }^{1} = h ⇒ (f ^{- }^{1})^{- }^{1} = h. It can be noted that h = f. Hence, (f ^{- }^{1})^{- }^{1} = f

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