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 by an example that $f*$ need not to be a function.
Attempt:
Let $A = B$ be the set of all real numbers greater or equal than $2$.
Define $f:A \to B$ by $f(x) = x^2+2x+3$ for all $x \in A$. Then, its inverse relation is $x = y^2 + 2y + 3$ for some $y \in B$. Note that \begin{align*} x &= y^2 + 2y+3 \\ y^2 + 2y + (3-x) &= 0 \\ y &= \frac{-2 \pm \sqrt{4-4(3-x)}}{2} \\ &= \frac{-2 \pm \sqrt{4x-8}}{2} \\ &= -1 \pm \sqrt{x-2} \end{align*} Hence, $f*(x) = -1 \pm \sqrt{x-2}$.
This has two values for a given value of $x$, so it is not a function; but it is still the inverse relation.
Is above true?