Function many to one Recently I come across this definition of function.
"A function F is a set of ordered pairs $( x , y)$, no two of which have
the same first member. That is, if $(x, y) \in F$ and $(x, z) \in F$, then $y = z$."
It made me confused because as far as i understood that it suggests we can't have one to many function like $\sqrt{x}$. But why this is so?
 A: A function $f\colon X \to Y$ can be thought of as a subset $F \subseteq X \times Y$ such that


*

*For every $x \in X$ there is a $y \in Y$ such that $(x, y) \in F$.

*If $(x, y), (x, y') \in F$ then $y = y'$.
If you have a function $f\colon X \to Y$ then it corresponds to the ordered pairs $F = \{(x, f(x)) \mid x \in X\}$.  Conversely if you have a set of pairs $F \subseteq X \times Y$ with the above properties then the associated function $f\colon X \to Y$ is defined by letting $f(x)$ be the unique element of $Y$ such that $(x, f(x)) \in F$.
It is indeed true that this means functions cannot be one to many.  In set theory this is part of the definition of a function.  A function $f\colon X \to Y$ is an assignment, to every $x \in X$, of a unique element $f(x) \in Y$.
A: The notion of function has been so successful in mathematics that it should not be tampered with.
Now it could very well be that in the realm of a certain problem you want to bring in some sort of "many-valued function". You can do this any time by making your "function" set-valued, or even multiset-valued. Then it is again a function in the official sense; but you would have to give explanations what "continuity" and the like would mean in your context.
An example: A polynomial
$$p(z):=z^n+a_{n-1}z^{n-1}+\ldots+a_1z+a_0$$
with complex coefficients $a_k$ has $n$ complex roots, counted with multiplicity. These roots depend on the coefficient vector ${\bf a}=(a_0,a_1,\ldots,a_{n-1})$, but there is no function $$f:\quad{\mathbb C}^n\to{\mathbb C},\qquad {\bf a}\to f({\bf a})$$ producing one or several of these roots upon inputting the coefficient vector ${\bf a}$. 
On the other hand you can define the set $M:={\mathbb N}^{\mathbb C}$ of multisets of complex numbers. Each such multiset $S$ is a function $S:\ {\mathbb C}\to{\mathbb N}$ which assigns to each complex number $z$ a certain integer multiplicity. Maybe $M$ is to large for our purpose; therefore we restrict to the set $M_n\subset M$  of multisets of cardinality $n$. Then the fundamental theorem of algebra tells us that there is a bona fide function
$$f:\quad {\mathbb C}^n\to M_n,\qquad {\bf a}\to f({\bf a})$$
producing a well-defined multiset of complex numbers upon inputting a coefficient vector ${\bf a}$. After defining a suitable topology on $M_n$ one can even say that the multiset $f({\bf a})$ depends continuously on ${\bf a}$.
