Why Is a Function Defined As Having Only One Y-Value Output?

Solution 1:

Strictly speaking, a "well-defined" function associates one, and only one, output to any particular input. A function would not be well-defined if say $y = f(x)$ such that $y$ can take on any number of values at any particular input. Suppose the input $\;x = a\;$ outputs more than one distinct value, so that $y = f(a) \in\{y_1, y_2, ..., y_k, ...\} $. We'd never be able to say what, precisely, $y$ is when $x = a$.

If $y = f(a) = b$, and $y = f(a) = c$, and $\,b\ne c$, then the value of the function $\,\,f(x) = y\,$ at $\,a\,$ is ambiguous: $y$ would not specify any particular value at $a$. That is, its value is not well-defined at $a$, and perhaps not well-defined at other inputs, as well. Nor can we say much about the behavior of a function at a particular value, if it can take on many values at a given point.

E.g. How would we define continuity of, say, a real-valued multifunction?

The strict definition of a function, in terms of "outputing" exactly one value for any given input is really no more than an attempt to keeping functions well-defined, and thus properties of functions well-defined.

But you'd might like to explore the following:

See this entry on multi-valued "functions":

A multivalued function (shortly: multifunction, other names: many-valued function, set-valued function, set-valued map, multi-valued map, multimap, correspondence, carrier) is a left-total relation; that is, every input is associated with at least one output.

The term "multivalued function" is, therefore, a misnomer because functions are single-valued. Multivalued functions often arise from functions which are not injective. Such functions do not have an inverse function, but they do have an inverse relation. The multivalued function corresponds to this inverse relation. [bold-face mine]

Solution 2:

Multi-valued functions do exist, but then that greatly limits our ability to say much about them. For functions, we're not just concerned about the value that it takes, but also the behavior of the function around the point. If we have multi-valued functions, then what we can say about the behavior of the point will be highly restricted, as it needs to be the same regardless of what value we choose.

A common example would be to let $f(x,y) = x^2 + y^2$, and ask for the values such that $f(x, y) = 1$ (i.e. the unit circle). The implicit function theorem is used, for us to conclude that there are 2 functions, namely $F(x) = \sqrt{1-y^2}$ and $F(x) = - \sqrt{1-y^2}$. If we had said that $F(x) = \pm \sqrt{1-y^2}$, it would be hard to talk about the behavior of $F'(x)$ exactly.