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Euclidean Geometry

Let f: A → B ...

Question

Let f : A → B and g : B → C be the bijective functions. Then, (gof)^–1 =
(a) f^–1o g^–1
(b) fog
(c) g^–1of^–1
(d) gof

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Solution

Given: f : A → B and g : B → C be the bijective functions

$Since, f : A \to B Thus, f^{- 1} : B \to A . . . (1) Since, g : B \to C Thus, g^{- 1} : C \to B . . . (2) From (1) and (2), we get f^{- 1} o g^{- 1} : C \to A . . . (3) Also, g o f : A \to C \Rightarrow {(g o f)}^{- 1} : C \to A . . . (4) Therefore, {(g o f)}^{- 1} = f^{- 1} o g^{- 1}$

Hence, the correct option is (a).

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Similar questions

f : A \to B a n d g : B \to A

be two functions such that

g o f = I_{A} a n f f o g = I_{B} .

Then, which of the following statement/s is/are true?

A = {1, 2, 3}, B = {α, β, λ} C = {p, q, r}

f : A \to B, g : B \to C

f = {(1, α), (2, λ) (3, β)}

g = {(α, q), (β, r), (λ, p)}

f

g

are bijective functions and

{(g o f)}^{- 1} = f^{- 1} o g^{- 1}

Q. Let A = {1, 2, 3, 4}; B = {3, 5, 7, 9}; C = {7, 23, 47, 79} and f : A → B, g : B → C be defined as f(x) = 2x + 1 and g(x) = x² − 2. Express (gof)⁻¹ and f⁻¹ og⁻¹ as the sets of ordered pairs and verify that (gof)⁻¹ = f⁻¹ og⁻¹.

A = {1, 2, 3, 4}

B = {3, 5, 7, 9}

C = {7, 23, 47, 79}

f : A \to B

f (x) = 2 x + 1

g : B \to C

g (x) = x^{2} - 2

{(g o f)}^{- 1}

f^{- 1} o g^{- 1}

in the form of ordered pair.

Q. Consider functions

f : A \to B

g : B \to C

(A, B, C \subseteq R)

(g o f)^{- 1}

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