I was studying mathematical analysis with the book Introduction to Mathematical Analysis (Parzynski and Zipse, 1987) and had a question regarding an exercise problem. It's actually the same one as this question but I was curious if my approach was also valid. The exercise is:
Show that if $f: X \rightarrow Y$ is one-to-one then $(f^{-1})^{-1} = f$.
The approach that I took is:
$$f: X \rightarrow Y\\f^{-1}: Y \rightarrow X \\ (f^{-1})^{-1} : X \rightarrow Y \\ \therefore (f^{-1})^{-1} = f$$
It seems a bit...insufficient and inconvincing.
The approach laid out in the question I linked above is:
Since we know that if $f$ is one-to-one, $f \circ f^{-1} = i_X$ and $f^{-1} \circ f = i_X$.
If we say that we have some function $g$ where $g \circ f^{-1} = i_X$, then $g = f$ since inverses are unique. Since the inverse of $f^{-1}$ is $g$, we can conclude that $(f^{-1})^{-1} = f$.
I was wondering if my approach also works, or if there should be a bit more rigor to it. Thanks.