# Show that $f^*:B \to A$ is a function iff $f$ is bijective as follows.

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?

$$\implies$$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)