Let $f: A \to B$. Let $f^*$ be the inverse relation, i.e. \begin{equation*} f^* = \{(y,x) \in B \times A \mid f(x)=y \}. \end{equation*} Show that $f^*:B \to $ is a function iff $f$ is bijective.
Attempt: Right now, I'm only have a problem for the right direction, but only on the injectivity proof.
Let $f^*$ is a function from $B$ to $A$. Let $x,z \in A$ and take some $y \in B$ such that $f(x) = f(z) = y$. Then, $(y,x),(y,z) \in f^*$. Since $f^*$ is a function, then we must have $x=z$. Hence, $f$ is injective.
Is the injective proof correct?
That looks fine, but you could also do this, which is a tad more elegant in my humble opinion:
$\implies$direction: Suppose that the inverse relation $f^* \subset B \times A$ is in fact a function $f^*: B \to A$. By definition, that means $f$ is invertible. But $f$ is invertible iff $f$ is bijective. Therefore $f$ is bijective. (See Theorem 1 in these notes from Northwestern: https://sites.math.northwestern.edu/~scanez/courses/300/notes/functions.pdf)