# Let $$f : X \to Y$$ be an invertible function. Show that the inverse of $f^{-1}$ is $f$, i.e., $(f^{-1})^{-1} = f$.

Toolbox:
• A function $g$ is called inverse of $f:x \to y$, then exists $g:y \to x$ such that $gof=I_x\;and\; fog=I_y$, where $I_x, I_y$ are identify functions.
• To prove inverse of $f^{-1}$ is f itself , we define a function $f^{-1}=g$ and show that $f^{-1}of = I_x$ and $f^{-1}off = I_y$ and hence $(f^{-1})^{-1}=f$
Given $f:X \to Y$ is invertible, therefore $f^{-1}$ exists
Let $f^{-1}=g$ such that $g:Y \to X$, and $gof=I_x\;and fog=I_y$
$\Rightarrow gof = I_x = f^{-1}of$
$\Rightarrow fog = I_y = fof^{-1}$
Therefore, $g = f^{-1}: Y \to X$ is invertible and f^{-1} is the inverse of $f$.
edited Mar 19, 2013