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Inverse of a Function

If f:A→ B ,...

Question

If $f : A \to B, g : B \to C$ and $h : C \to D$ are bijectives then $(h o g o f)^{- 1}$ is

A

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

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B

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

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C

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

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D

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

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Solution

The correct option is B $f^{- 1} o g^{- 1} o h^{- 1}$
Given $f, g, h$ are bijective mappings.
So $f, g, h$ all are invertible.
We know,
$(f o g)^{- 1} = g^{- 1} o f^{- 1}$
Consider , $(h o g o f)^{- 1}$

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

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

$= f^{- 1} o g^{- 1} o h^{- 1}$

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Similar questions

Q. 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

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?

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}, 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}

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.

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