Let $f : X \to Y$ be an invertible function. Show that $f$ has unique inverse

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Solution

$L e t f : X \to Y b e a n i n v e r t i b l e f u n c t i o n .$

If g is an inverse of $f,$ then for all $y ϵ Y$ ;
$f o g (y) = I_{y}$

Let $g 1$ and $g 2$ be two inverses of $f,$

Then,for all
$y ϵ Y, f o g 1 (y) = I_{y} a n d f o g 2 (y) = I_{y} \Rightarrow f o g 1 (y) = f o g 2 (y) f (g 1 (y)) = f (g 2 (y)) - (1)$

Since f is invertible.

f is one-one.

$I f f (x 1) = f (x 2), t h e n x 1 = x 2 S i n c e, f (g 1 (y)) = f (g 2 (y)), ∴ g 1 (y) = g 2 (y) \Rightarrow g 1 = g 2$

hence f has a unique inverse.

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