Surjection iff Image equals Codomain
Contents |
Theorem
A mapping $f: S \to T$ is a surjection iff its image equals its codomain:
- $\operatorname{Im} \left({f}\right) = \operatorname{Cdm} \left({f}\right)$
That is, iff $f \left({S}\right) = T$.
Some sources use this as the definition of a surjection.
Proof
Let $f: S \to T$ be a mapping.
Necessary Condition
First suppose that $\operatorname{Im} \left({f}\right) = T$.
Then by the definition of set equality, $T \subseteq \operatorname{Im} \left({f}\right)$.
Hence:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle y\) | \(\in\) | \(\displaystyle T\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle y\) | \(\in\) | \(\displaystyle \operatorname{Im} \left({f}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
So by the definition of the image of $f$, $\exists x \in \operatorname{Dom} \left({f}\right): f \left({x}\right) = y$.
Thus $f$ is a surjection.
$\Box$
Sufficient Condition
Now suppose that $f: S \to T$ is a surjection.
We already have $\operatorname{Im} \left({f}\right) \subseteq T$ from Image is Subset of Codomain.
So all we need to do is prove that $T \subseteq \operatorname{Im} \left({f}\right)$.
Thus:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle y\) | \(\in\) | \(\displaystyle T\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle \exists x \in \operatorname{Dom} \left({f}\right): f \left({x}\right)\) | \(=\) | \(\displaystyle y\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of Surjection | ||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle y\) | \(\in\) | \(\displaystyle \operatorname{Im} \left({f}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of image of $f$ |
By the definition of a subset, this proves that $T \subseteq \operatorname{Im} \left({f}\right)$.
Thus by the definition of set equality, $T = \operatorname{Im} \left({f}\right)$ and the proof is complete.
$\blacksquare$
Sources
- Iain T. Adamson: Introduction to Field Theory (1964)... (previous)... (next): Chapter $1 \ \S 3$
- T.S. Blyth: Set Theory and Abstract Algebra (1975): $\S 5$
- Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 21$
- H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability (1996): Appendix $\text{A}.5$: Proposition $\text{A}.5.1: 2 \ \text{(b, c)}$