What is the Definition and Understanding of Surjective Functions?

[Phys12]

In my book, the definition of surjection is given as follows:

Let A and B be sets and f:A->B. The function f is said to be onto if, for each b ϵB, there is at least one a ϵ A for which f(a)=b. In other words, f is onto if R(f)=B. A function which is onto is also called a surjection or a surjective function.

However, what I don't understand is why does there need to be at least one a ϵ A? Shouldn't there be only one since it's a function and a function by definition, for a given image, cannot have 2 pre-images?

 

[fresh_42]

In my book, the definition of surjection is given as follows:

Let A and B be sets and f:A->B. The function f is said to be onto if, for each b ϵB, there is at least one a ϵ A for which f(a)=b. In other words, f is onto if R(f)=B. A function which is onto is also called a surjection or a surjective function.

However, what I don't understand is why does there need to be at least one a ϵ A? Shouldn't there be only one since it's a function and a function by definition, for a given image, cannot have 2 pre-images?

No. A function cannot have two $b \in B$ for the same $a \in A$. It can, however, have two elements $a$ which map onto the same element $b$. E.g. $f\, : \,x \longmapsto x^2$ is a function, and $f(-1) = f(+1)$. The relation $x \longmapsto \pm \sqrt{x}$ is no function, only if we restrict ourselves to either $+\sqrt{x}$ or $-\sqrt{x}$, but not both. $f \, : \,\mathbb{R} \longrightarrow \mathbb{R}$ with $f(x)=x^2$ is not surjective, because the range is only $\mathbb{R}_0^+ \subsetneq \mathbb{R}$. But $f \, : \, \mathbb{R} \longrightarrow \mathbb{R}_0^+$ with $f(x)=x^2$ is surjective.