prove that $f^{-1} = f*$ as follow edit:
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 if $f* : B \to A$ is a function,
then $f^{-1} = f*$.
Attempt:
Let $f*: B \to A$ be a function.
Then, $f$ is bijective and hence $f$ is invertible, i.e. $f$ have an inverse, say $f^{-1}$.
It is clear that $f^{-1} \circ f = i_A$
and $f \circ f* = i_B$ where $i$ is the identity function. Hence,
\begin{equation*}
f^{-1} = f^{-1} \circ i_B = f^{-1} \circ (f \circ f*) = (f^{-1} \circ f) \circ f* = i_A \circ f* = f*.
\end{equation*}
Thus, $f^{-1} = f$.
Is the above correct?
First, suppose that $f*$ is a function. That means that each for each $y \in B, \exists!x \in A$ such that $f^*(y)=x.$ Then $(f^* \circ f)(x) = f^*(f(x)) = f^*(y)=x,$ and similarly, $(f \circ f^*)(y) =f(f^*(y))= f(x)=y$, so $f*$ is the inverse of $f$. Or perhaps I should say $f*$ is an inverse of $f$, since you've also been tasked with proving the uniqueness of the inverse.
To prove that the inverse of f is unique, suppose that f has two inverses, g and h. We need to show that g=h. Since g is the inverse of f, that means $g∘f=id_A$  and $f∘h=id_B$. Then $g=g∘id_B=g∘(f∘h)=(g∘f)∘h=id_A∘h=h$.
First, suppose that $f*$ is a function. That means that each for each $y \in B, \exists!x \in A$ such that $f^*(y)=x.$ Then $(f^* \circ f)(x) = f^*(f(x)) = f^*(y)=x,$ and similarly, $(f \circ f^*)(y) =f(f^*(y))= f(x)=y$, so $f*$ is the inverse of $f$. Or perhaps I should say $f*$ is an inverse of $f$, since you've also been tasked with proving the uniqueness of the inverse.
To prove that the inverse of f is unique, suppose that f has two inverses, g and h. We need to show that g=h. Since g is the inverse of f, that means $g∘f=id_A$  and $f∘h=id_B$. Then $g=g∘id_B=g∘(f∘h)=(g∘f)∘h=id_A∘h=h$.