Q. 34.0( 4 Votes )

Consider f : {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-1og-1.


We know (f-1og-1)(x) = f-1(g-1(x))


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


(f-1og-1)(apple) = f-1(g-1(apple)) = f-1(a) = 1


(f-1og-1)(ball) = f-1(g-1(ball)) = f-1(b) = 2


(f-1og-1)(cat) = f-1(g-1(cat)) = f-1(c) = 3


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


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


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