Surjection iff Image equals Codomain

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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.

$\blacksquare$


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