Theorem. If $f: X \to Y$ and $g: Y \to X$ are bijective maps, then the following assertions hold:
(1) $f^{-1}$ is bijective with $(f^{-1})^{-1}=f$
Proof is left as an exercise.
Things we already know (stated in the book)
Definition 2.18: Let $f: X \to Y$ and $g: Y \to Z$. Thus, $g \circ f: X \to Z$
Definition 2.21: $f = g^{-1}$ and $g = f^{-1}$
My attempt:
$(f^{-1})^{-1}=f$
$g^{-1}=f$
$f = f$
Is the proof correct? Seems so simple it makes me nervous, haha.
Liesen, J., Mehrmann, V. 2015. Linear Algebra. Berlin, Germany.: Springer.
Your proof is not correct, for three reasons:
Your proof is kind of written "backwards". A proof should always be a finite sequence of logical steps, starting from what you already know to be true and ending with what you wanted to prove. In this problem, we already know that $f$ is a function, so we can start by saying $f = f$. It's not valid to start by saying $(f^{-1})^{-1} = f$, since this is part of what we want to prove!
"Things we already know: $f=g^{−1}$ and $g=f^{−1}$": this is incorrect! There is no assumed relationship at all between $f$ and $g$ in the statement of the theorem. You can be absolutely sure of this because if $f = g^{-1}$ were true, then we would have $g : Y \to X$, but instead the assumption is $g : Y \to Z$ (it seems you typo'd this when copying down the theorem statement), and $Z$ is not neccesarily equal to $X$. I think you're getting this from Definition 2.21, but you need to be careful about this sort of thing in math: the $f$ in that definition is an arbitrary bijection, and the $f$ in this theorem is another arbitrary bijection. There is absolutely no reason that these two $f$'s have to be the same function – indeed neither one of them is any function at all in particular.
You actually need to prove two separate things here: first, that $f^{-1}$ is bijective, and second that the inverse of $f^{-1}$ is $f$ (that's what is meant by "$f^{-1}$ is bijective with $(f^{-1})^{-1} = f$").