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

Given, 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 = X and fog = 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's 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 = X and hog = Y, where X = {1, 2, 3} and Y = {a, b, c}.
⇒ The inverse of g exists and g⁻¹= h
⇒ (f⁻¹)⁻¹ = h
therefore, h = f
hence,  (f⁻¹)⁻¹ = h

Answer 2

Answer:

Given, 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 = X and fog = 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's 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 = X and hog = Y, where X = {1, 2, 3} and Y = {a, b, c}.

⇒ The inverse of g exists and g⁻¹= h

⇒ (f⁻¹)⁻¹ = h

therefore, h = f

hence,  (f⁻¹)⁻¹ = h

Step-by-step explanation: