An ordered pair drawn from ℝ² is in some sense a single value. In another sense, it is two values. Which way you want to think about it depends on what you're going to do with it; if you're thinking about square roots of real numbers, it will be more useful to think of it as two values.
> It seems the parent has just mixed up the domain and codomain, because under that assumption he would be right about both the definition and bijectivity
He still wouldn't be right about bijectivity; the condition that an element of the codomain can have no more than one inverse only describes an injective function.
What I tryed to say is that the sense in which “exactly one element” is used in the definition of function is inclusive of codomain being R^n, so it confused me why you would provide a function that has a codomain of R^2 as something that suggests deviation from the formalism. It just seemed misleading to phrase it that way
I thought consistently would convey the idea I had, nevermind if it doesn’t
My bad about bijectivity, I see it now, you’re right
Isn’t the function definition meant to be interpreted as “maps consistently to exactly one element of the codomain”? So an ordered pair of R^2 is still one element of R^2
It seems the parent has just mixed up the domain and codomain, because under that assumption he would be right about both the definition and bijectivity