Question

Let f: X → Y be an invertible function. Show that f has unique inverse. (Hint: suppose g1 and g2 are two inverses of f. Then for all y ∈ Y, fog1(y) = IY(y) = fog2(y). Use one-one ness of f).

Answer 1

given, f: X → Y be an invertible function.
Also, suppose f has two inverse
Then, for all y ∈ Y, we get:
fog₁(y) = I₁ (y) = fog₂(y)
=> f(g₁(y)) = f(g₂(y))
=> g₁(y) = g₂(y). [if f is inversible
then , f is definitely one one function.
it means, if f(x) =f (y) then, x = y.]
=> g₁ = g₂
$\textbf{Therefore, f has a unique inverse.}$

Answer 2

Answer:

Step-by-step explanation:

given, f: X → Y be an invertible function.

Also, suppose f has two inverse

Then, for all y ∈ Y, we get:

fog₁(y) = I₁ (y) = fog₂(y)

=> f(g₁(y)) = f(g₂(y))

=> g₁(y) = g₂(y). [if f is inversible

then , f is definitely one one function.

it means, if f(x) =f (y) then, x = y.]

=> g₁ = g₂