Consider f: {1, 2, 3} → {a, b, c} given by f(1) = a, f(2) = b and f(3) = c. Find f^–1 and show that (f^–1)^–1 = f.

It is given that f: {1, 2, 3} → {a, b, c} given by

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

So, if we define g:

{a,b,c} → {1, 2, 3} as

g(a) = 1, g(b) = 2, g(c) = 3, then we get:

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

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

Then, f^-1: {a, b, c} → {1, 2, 3} is given by

f^-1(a) = 1, f^-1(b) = 2, f^-1(c) = 3

Let us now find the inverse of f-1,

So, if we define h: {1, 2, 3} → {a, b, c} as

h(1) = a, h(2) = b, h(3) = c, then we get:

(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}.

⇒ The inverse of g exists and g^-1 = h

⇒ (f^-1)^-1 = h

⇒ h = f

⇒ (f^-1)^-1 = f