1.

Let f : A → B, g : B → C be bijections. Then prove that (gof)–1 = f–1og–1.

Answer»

f : A → B, g : B → C are bijections

⇒ gof : A → C is a bijection

Also g–1 : C → B and f–1 : B → A are bijections

⇒ f –1og–1 : C → A is a bijection.

Let c be any element of C.

Then ∃ an element b ∈ B such that g(b) = c

⇒ b = g–1(c)

Also ∃ an element a ∈ A such that f(a) = b

⇒ a = f–1 (b)

Now (gof) (a) = g(f(a)) = g(b) = c

⇒ a = (gof)–1 (c) ⇒ (gof)–1 (c) = a ——— (1)

Also (f–1og–1) (c) = f–1(g–1(c)) =f–1(b) = a ——— (2)

∴ From (1) and (2); (gof)–1(c)

= (f–1og–1) ( c)

⇒ (gof)–1 = f–1og–1.


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