Question

If f : A
→ B, g: B
→ C be bijections, then show that :
(gof)-1 = -log-1.​

Answer 1

Step-by-step explanation:

this is brainliest answer

Answer 2

Final answer: (gof)⁻¹ = f⁻¹og⁻¹

Given that: We are given f : A → B, g: B → C be bijections.

To find: We have to find (gof)⁻¹ = f⁻¹og⁻¹

Explanation:

• A function f: X → Y is said to be bijective if f is satisfy both injective and surjective function properties, which means function f satisfy both one-one and onto.

i.e., every element “x” in the codomain X, there is exactly one element “y” in the domain Y.

• f: A → B is a bijection.

• g: B → C is a bijection.

• (gof)⁻¹: C → A is a bijection.

as f: A → B, g: B → C  are bijections.

• g⁻¹: C → B, f⁻¹: B → A are also bijections.

• f⁻¹og⁻¹: C→A is a bijection.

• (gof)⁻¹, f⁻¹og⁻¹ have same domain and codomain.

as f: A → B is onto

if b ϵ B then a ϵ A such that

f(a) = b

f⁻¹(b) = a

as g: B → C is onto

if c ϵ C then  b ϵ B such that

g(b) = c

g⁻¹(c) = b

• (gof)(a) = g(f(a)) = g(b) = c

(gof)(a) = c = (gof)⁻¹(c) = a

f⁻¹og⁻¹(c) = f⁻¹(g⁻¹(c)) = f⁻¹(b) = a

f⁻¹og⁻¹(c) = a

• f⁻¹og⁻¹=(gof)⁻¹

Hence proved.

To know more about the concept please go through the links

https://brainly.in/question/17194321

https://brainly.in/question/7026829

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