Theorem: Let both $f$ and $f^{-1}$ be functions. $\newcommand{\dom}{\operatorname{dom}}\newcommand{\im}{\operatorname{im}}$ Then $\dom(f) = \im(f^{-1})$ and $\dom(f^{-1}) = \im(f)$.
Let $f: X \to Y$ be an injective function. It might not be true that $f^{-1}: Y \to X$ since it's not necessarily true that $f(X) = Y$. So for $f^{-1}$ to exist $f$ must also be surjective, right? So, suppose that $f$ is a non-surjective function. Does it contradict the theorem above since according to it $f$ is any function? I know that there's a hole somewhere in my reasoning, but I can't seem to find it.