# 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 $$y \in Y$$, $$fog_1(y) = 1 _Y (y) = fog_2 (y).$$   Use ono-one ness of $$f$$)

Toolbox:
• To show that $f:X \to Y$ has unique inverse function we take two functions $g_1\; and \;g_2$ as two inverse functions of f then show $g_1=g_2$
Given $f$ is invertible. We need to show that $f$ has a unique inverse.
If a function $f$ has two inverses $g_1$ and $,g_2$ then $fog_1 (y)=I_y\;and\;fog_2(y)=I_y$ for $y \in Y$
$\Rightarrow fog_1 (y)=I_y(y)=fog_2(y)$
$\Rightarrow f(g_1(y))=f(g_2(y))$
Since $f$ is invertible $f$ is one-one, $\Rightarrow g_1(y)=g_2(y)$
There if $f$ has two inverses, then $g_1=g_2$. In otherwords, $f$ will only have a unique inverse.
edited Mar 19, 2013