If $f: A \to B$ is bijective show that $f^{-1}$ is unique.
Suppose that $f^{-1}$ and $h$ are inverses of $f$. Let $y = f(x)$. We need to show that $f^{-1}(y) = h(y)$ for all $y \in B$. So $f^{-1}(f(x)) = h(f(x)) = x$ for all $y \in B$, thus $f^{-1}$ is unique?
If $f: A \to B$ is bijective show that $f^{-1}$ is unique?
Let $g,h: B\mapsto A$ such that $g(f(a)) =h(f(a))= a$ for all $a\in A$. To show that $g=h$ we need to show to things, (i) they have the same domain (ii) $g(x)=h(x)$ for all $x$ in domain. The first part is straightforward, it is given that $g,h$ have domain $B$. Now let $x\in B$. Then since $f$ is bijective it means $x=f(a)$ for some $a\in A$. Thus, $g(x) = g(f(a)) = a = h(f(a)) = h(x)$. And thus these are the same functors.