Surjection/Examples/Non-Surjection/Arbitrary Mapping on Sets

Example of Mapping which is Not a Surjection

Let $A = \set {a, b, c}$.

Let $B = \set {1, 2, 3}$.

Let $f \subseteq {A \times B}$ be the mapping defined as:

$f = \set {\tuple {a, 2}, \tuple {b, 1}, \tuple {c, 1} }$

Then $f$ is not a surjection.

Proof

For $f$ to be a surjection, it would be necessary that:

$\forall y \in B: \exists x \in A: \map f x = y$

However $3 \in B$ does not fit this condition.

Hence $f$ is not a surjection.

$\blacksquare$

Sources

• 1977: Gary Chartrand: Introductory Graph Theory ... (previous) ... (next): Appendix $\text{A}.4$: Functions