Show that $ (f^{-1})^{-1}=f $ $$
f:X\to Y 
$$
$f$ is invertible, show that $(f^{-1})^{-1}=f$
Here it is not given that how the function is defined, so I think that making equations and solving them will not help me.
So I have given some arguments that $f^{-1}:Y\to X$ and is defined in the inverse manner of that of $f$.
Now $(f^{-1})^{-1}:X\to Y$ and it will be defined in the inverse manner of that of $f^{-1}$, we know that inverse of a function is unique.
Since $f^{-1}$ is the inverse of $f$ and $(f^{-1})^{-1}$ is the inverse of $f^{-1}$
So, 
$$
(f^{-1})^{-1}=f
$$ 
I'm not totally satisfied with my solution so is there any other way to solve this problem.
 A: Assume $(f^{-1})^{-1}=g \ $ where $g$ is some function. Therefore, you have $gf^{-1}=f^{-1}g=e \ $ where $e$ is the identity function such that for any function $h$, $eh = he = h$.
Now, you can do:
$$ f^{-1}g = e \\ ff^{-1}g = fe \\ g = f$$
There you have it.
A: Hint: per definition you know that $(f^{-1})^{-1}\circ f^{-1}=\mathrm{id}_Y$.
A: We want to prove that $(f^{-1})^{-1}(x)=f(x) $
Let $f$ be a bijection such that $f:A\to B$ which is one to one and onto and so $f^{-1}$ exists therefore becomes $f^{-1}:B\to A$ where $f^{-1}$ is also a bijection i.e one to one and onto hence invertible and thus it's inverse exists $(f^{-1})^{-1}$ and now $(f^{-1})^{-1}:A\to B$ where it's located a bijection i.e one to one and onto.
If $x\in A$ the $D(f)= D((f^{-1})^{-1})$
Let $f(x)= y$ then $f^{-1}(y)=x$
We can now see clearly that $(f^{-1})^{-1}(x)= y$ and since $f(x)= y$  and by equality of function property $f(x)=y=(f^{-1})^{-1})(x)$
A: how about this?
$f(a)=b$
$f$ is invertible, it is one-to-one correspondence.
$f^{-1}(b)=a$
$(f^{-1})^{-1}(a)=b=f(a)$
therefor, $(f^{-1})^{-1}=f$
