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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?

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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)

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    $\begingroup$ okay noted. thanks! $\endgroup$
    – lap lapan
    Jan 6, 2021 at 7:57
  • $\begingroup$ @user795084 You're welcome, and if you like my answer, please feel free to select it as the official answer. $\endgroup$
    – Hank Igoe
    Jan 6, 2021 at 8:00
  • $\begingroup$ Can you help me answer my question before this? $\endgroup$
    – lap lapan
    Jan 6, 2021 at 8:01
  • $\begingroup$ @user795084 Sure, do you mean in the other direction? $\endgroup$
    – Hank Igoe
    Jan 6, 2021 at 8:02
  • $\begingroup$ yes! the question about uniqueness of inverse $\endgroup$
    – lap lapan
    Jan 6, 2021 at 8:03

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