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

(Hint : suppose ${g_1}$ and ${g_2}$ are two inverses of f. Then for all $y \in Y,fo{g_1}(y) = {I_Y}(y) = fo{g_2}(y).$ Use one-one ness of f).

Given f : $X \to Y$be invertible.

Thus, f is one-one and onto and therefore ${f^{ - 1}}$ exists.

Let ${g_1}$and g2 be the two inverses of f. Now for all $y \in Y,$ $fo{g_1}(y) = {I_Y}(y) = fo{g_2}(y)$

$\Rightarrow$ $fo{g_1}(y) = fo{g_2}(y) \Rightarrow f[{g_1}(y)] = f[{g_2}(y)] \Rightarrow {g_1}(y) = {g_2}(y)$

Hence, f has a unique inverse.