Why is one relation a function and the other not?

Question

Why does $\{(x, x^2) \mid x \in\mathbb{R}\}$ qualify as a function, whereas $\{(x^2, x) \mid x\in\mathbb{R}\}$ does not?

Answer 1

The second cannot define a function because what would be the image of $9$?

$$9=(3)^2=(-3)^2.$$

$3$ or $-3$ ?

It would be a function if the domain was $\Bbb R^+$ and the codomain was either $\Bbb R^+$ or $\Bbb R^-$.

For example, in the latter case, we will have $f (16)=-4$.

Answer 2

Note that $(1,-1)$ and $(1,1)$ both belong to the second relation. This violates the condition of being a function since the first coordinates are the same, but the second coordinates are not. I.e., $1$ is not mapped to a well-defined value. In fact, no number except $0$ is.

Answer 3

A relation on $X$ is a collection of ordered pairs in the Cartesian product $X\times Y$. There are no other requirements---any collection of ordered pairs is a relation.

A function is a special kind of relation $R$ where we impose the additional requirement that if $(x,y) \in R$ and $(x,y')\in R$, then it must be the case that $y = y'$. That is, any element of the domain $X$ is paired with exactly one element in the codomain $Y$.

In your example, $\{ (x,x^2) : x \in \mathbb{R} \}$ is a relation, since we have $$ y' = x^2 = y \iff y = y'.$$ On the other hnd, $\{ (x^2,x) : x\in\mathbb{R} \} =: R$ is not a function, since both $(4,2)$ and $(4,-2)$ are in this relation. Thus we have $(4,2), (4,-2) \in R$, but $2 \ne -2$.