Q. 34.0( 4 Votes )

# Consider f : {1, 2, 3} → {a, b, c} and g : {a, b, c} → {apple, ball, cat} defined as f(1) = a, f(2) = b, f(3) = c, g(a) = apple, g(b) = ball and g(c) = cat. Show that f, g and gof are invertible. Find f^{-1}, g^{-1}, (gof)^{-1} and show that (gof)^{-1} = f^{-1}og^{-1}.

Answer :

f : {1, 2, 3} → {a, b, c} and f(1) = a, f(2) = b, f(3) = c

⇒ f = {(1, a), (2, b), (3, c)}

Recall that a function is invertible only when it is both one-one and onto.

Here, observe that distinct elements of the domain {1, 2, 3} are mapped to distinct elements of the co-domain {a, b, c}.

Hence, f is one-one.

Also, each element of the range {a, b, c} is the image of some element of {1, 2, 3}.

Hence, f is also onto.

Thus, the function f has an inverse.

We have f^{-1} = {(a, 1), (b, 2), (c, 3)}

g : {a, b, c} → {apple, ball, cat} and g(a) = apple, g(b) = ball, g(c) = cat

⇒ g = {(a, apple), (b, ball), (c, cat)}

Similar to the function f, g is also one-one and onto.

Thus, the function g has an inverse.

We have g^{-1} = {(apple, a), (ball, b), (cat, c)}

We know (gof)(x) = g(f(x))

Thus, gof : {1, 2, 3} → {apple, ball, cat} and

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

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

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

⇒ gof = {(1, apple), (2, ball), (3, cat)}

As the functions f and g, gof is also both one-one and onto.

Thus, the function gof has an inverse.

We have (gof)^{-1} = {(apple, 1), (ball, 2), (cat, 3)}

Now, let us consider f^{-1}og^{-1}.

We know (f^{-1}og^{-1})(x) = f^{-1}(g^{-1}(x))

Thus, f^{-1}og^{-1} : {apple, ball, cat} → {1, 2, 3} and

(f^{-1}og^{-1})(apple) = f^{-1}(g^{-1}(apple)) = f^{-1}(a) = 1

(f^{-1}og^{-1})(ball) = f^{-1}(g^{-1}(ball)) = f^{-1}(b) = 2

(f^{-1}og^{-1})(cat) = f^{-1}(g^{-1}(cat)) = f^{-1}(c) = 3

⇒ f^{-1}og^{-1} = {(apple, 1), (ball, 2), (cat, 3)}

Therefore, we have (gof)^{-1} = f^{-1}og^{-1}.

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