Let $f: X \rightarrow Y$ be an invertible function. Show that $f$ has a unique inverse.
(Hint: Suppose $g_{1}$ and $g_{2}$ are two inverses of $f$. Then for all $y \in Y$,$f \circ g_{1}(y) = I_{Y}(y) = f \circ g_{2}(y)$. Use the one-one property of $f$.)

Let $f: X \rightarrow Y$ be an invertible function.
Suppose $f$ has two inverses,$g_{1}$ and $g_{2}$,where $g_{1}: Y \rightarrow X$ and $g_{2}: Y \rightarrow X$.
By the definition of an inverse function,for any $y \in Y$,we have:
$f \circ g_{1}(y) = I_{Y}(y) = y$
$f \circ g_{2}(y) = I_{Y}(y) = y$
Therefore,$f(g_{1}(y)) = f(g_{2}(y))$ for all $y \in Y$.
Since $f$ is an invertible function,it must be a bijection (both one-one and onto).
Because $f$ is one-one,$f(x_{1}) = f(x_{2}) \Rightarrow x_{1} = x_{2}$.
Applying this to $f(g_{1}(y)) = f(g_{2}(y))$,we get $g_{1}(y) = g_{2}(y)$ for all $y \in Y$.
Thus,$g_{1} = g_{2}$.
Hence,the inverse of an invertible function $f$ is unique.