Question

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.

Answer 1

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

f(1) = a, f(2) = b and f(3) = c

If we define g: {a, b, c} → {1, 2, 3} as g(a) = 1, g(b) = 2, g(c) = 3, then we have:

(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

∴ 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.

∴ 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 i.e., find the inverse of g.

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

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

(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}.

Thus, the inverse of g exists and g⁻¹ = h ⇒ (f⁻¹)⁻¹ = h.

It can be noted that h = f.

Hence, (f⁻¹)⁻¹ = f.