Functions:Surjective

A function is surjective, or "onto", if for all $b\in B$, there exists at least one $a\in A$ such that $f(a)=b$; that is, every element in the codomain is mapped to by at least one element in the domain. Another way to state this is that a function is surjective if and only if its range (all outputs of the function) equals its codomain.

Examples:

• Let $A=\{-2,-1,0,1,2\}$ and $B=\{1,3,9\}$, and let $f:A\to B$ such that $f(-2)=9$, $f(-1)=3$, $f(0)=1$, $f(1)=3$, and $f(2)=9$. Each element of $B$ is mapped to by at least one element of $A$, so $f$ is surjective.

• For the same $A$ and $B$ as in the previous example, let $g:A\to B$ be a function such that $g(-2)=1$, $g(-1)=1$, $g(0)=1$, $g(1)=1$, and $g(2)=1$. This is not a surjective function since there are elements in the codomain that are not mapped to by any elements of the domain.

• Let $f:\mathbb {R} \to \mathbb {R} ,f(x)=x^{3}$. For every $y\in \mathbb {R}$, there exists $x={\sqrt[{3}]{y}}\in \mathbb {R}$ such that $f(x)=({\sqrt[{3}]{y}})^{3}=y$. So, $f$ is a surjective function.

• Let $g:\mathbb {R} \to \mathbb {R} ,g(x)={\frac {1}{x}}$. $0$ is in the codomain, but there is no $x\in \mathbb {R}$ such that ${\frac {1}{x}}=0$. Thus, $g$ is not surjective.

Resources

• Course Textbook, pages 154-164