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.