# Functions and mappings?

My textbook says that there is Cartesian product $A \times B$ which produces yet another set of all possible pairs $(a \in A, b \in B)$. So far so good. Than something really unexpected happens:

Subsets of $A \times B$ are called 'relations'. We will define a mapping or a function $f \subset A \times B$ from set $A$ to set $B$ to be the special type of relation in which for each element $a \in A$ there is a UNIQUE element $b \in B$ such that $(a, b) \in f$.

Seems like $f(x) = x^2$ is not a function anymore nor a mapping? Since any positive output "is mapped" with two rather than strictly one input, i.e. $2^2 = 4, -2^2 = 4$.

Wouldn't it be more correct to define mapping as injective function also known as "one-to-one" function? So "maps" are more strict kind of functions?

• But notice that even in your example, $f(x)=x^2$, for each $x$ there is only one element $y$ with $(x,y) \in f$, namely, $y=x^2$. You are confusing things... – amrsa Jan 20 '18 at 20:13
• You're thinking uniquely about the wrong number. For each $x$ there is a unique $x^2$. For $a=1$, the unique $b$ is $1$, and for $a=-1$ the unique $b$ is also $1$. For each $a=x$, the unique $b$ is what we call $f(x)$. – Arthur Jan 20 '18 at 20:14

Mappings and functions are the same.

By "unique $b$", the definition is saying that $a$ appears first in exactly one tuple $(a,b)$ (i.e. each $a$ doesn't have two different $b$'s $b_1$ and $b_2$ for which $(a, b_1)$ and $(a, b_2)$ are both in the relation).

It does not necessarily mean that for any $a_1$ and $a_2$, if $(a_1, b_1)$ and $(a_2, b_2)$ are in the relation, then either $a_1=a_2$ or $b_1 \neq b_2$. (In other words, it is okay for each $b$ to correspond to more than one $a$).

In the example of $f(x)=x^2$, some outputs, or $b$'s, have more than one $a$ mapped to them, but each input, or $a$, has exactly one $b$ to which it is mapped.