Show that $ (f^{-1})^{-1}=f $

Question

$$ f:X\to Y $$ $f$ 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.

Answer 1

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.

Answer 2

Hint: per definition you know that $(f^{-1})^{-1}\circ f^{-1}=\mathrm{id}_Y$.

Answer 3

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, $(f^{-1})^{-1}=f$