Question

Consider f: {1, 2, 3} \(\to\) {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} \(\to\) {a, b, c} is given by, 
f(1) = a, f(2) = b, and f(3) = c 
If we define g: {a, b, c} \(\to\) {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=IY , 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} \(\to\) {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} \(\to\) {a, b, c} as 
h(1) = a, h(2) = b, h(3) = c, then we have:
(gof(1)) = g(f(1)) = g(a) = 1 
(gof(2)) = g(f(2)) = g(b) = 2 
(gof(3)) = g(f(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, hog=I_y where X = {1, 2, 3} and Y = {a, b, c}. 
Thus, the inverse of g exists and g⁻¹ = h
\(\implies\) (f⁻¹)⁻¹ = h.
It can be noted that h = f. 

Hence, (f⁻¹)⁻¹ = f.