Page 17 - Relations and Functions 19.10.06.pmd

RELATIONS AND FUNCTIONS 17

Solution Note that by definition, f and g are bijective functions. Let
f : {a, b, c} → (1, 2, 3} and g : {apple, ball, cat} → {a, b, c} be defined as
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f {a} = 1, f {b} = 2, f {c} = 3, g {apple} = a, g {ball} = b and g {cat} = c.
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It is easy to verify that f of = I {1, 2, 3} , f o f = I {a, b, c} , g og = I {a, b, c} and go g = I ,
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D
where, D = {apple, ball, cat}. Now, gof : {1, 2, 3} → {apple, ball, cat} is given by
gof(1) = apple, gof(2) = ball, gof(3) = cat. We can define
(gof) : {apple, ball, cat} → {1, 2, 3} by (gof) (apple) = 1,(gof) (ball) = 2 and
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(gof) (cat) = 3. It is easy to see that (g o f) o (g o f) = I {1, 2, 3} and
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(gof) o (gof) = I . Thus, we have seen that f, g and gof are invertible.
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D
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Now, f og (apple)= f (g (apple)) = f (a) = 1 = (gof) (apple)
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f og (ball) = f (g (ball)) = f (b) = 2 = (gof) (ball) and
f og (cat) = f (g (cat)) = f (c) = 3 = (gof) (cat).
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Hence (gof) = f og .
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The above result is true in general situation also.
Theorem 2 Let f : X → Y and g : Y → Z be two invertible functions. Then gof is also
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invertible with (gof) = f og .
Proof To show that gof is invertible with (gof) = f og , it is enough to show that
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(f og )o(gof) = I and (gof)o(f og ) = I .
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X Z
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Now, (f og )o(gof)= ((f og ) og) of, by Theorem 1
= (f o(g og)) of, by Theorem 1
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= (f oI ) of, by definition of g –1
Y
= I .
X
Similarly, it can be shown that (gof )o (f og ) = I .
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Z
Example 28 Let S = {1, 2, 3}. Determine whether the functions f : S → S defined as
below have inverses. Find f , if it exists.
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(a) f = {(1, 1), (2, 2), (3, 3)}
(b) f = {(1, 2), (2, 1), (3, 1)}
(c) f = {(1, 3), (3, 2), (2, 1)}
Solution
(a) It is easy to see that f is one-one and onto, so that f is invertible with the inverse
f of f given by f = {(1, 1), (2, 2), (3, 3)} = f.
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(b) Since f (2) = f (3) = 1, f is not one-one, so that f is not invertible.
(c) It is easy to see that f is one-one and onto, so that f is invertible with
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f = {(3, 1), (2, 3), (1, 2)}.