Equivalence of Definitions of Surjection

Theorem

The following definitions of the concept of Surjection are equivalent:

Definition 1

$f: S \to T$ is a surjection if and only if:

$\forall y \in T: \exists x \in \Dom f: \map f x = y$

That is, if and only if $f$ is right-total.

Definition 2

$f: S \to T$ is a surjection if and only if:

$f \sqbrk S = T$

or, in the language and notation of direct image mappings:

$\map {f^\to} S = T$

That is, $f$ is a surjection if and only if its image equals its codomain:

$\Img f = \Cdm f$

Proof

Definition 1 implies Definition 2

Let $f$ be a mapping which fulfils the condition:

$\forall y \in T: \exists x \in \Dom f: f \paren x = y$
$\Img f \subseteq T$

It remains to be proved that:

$T \subseteq \Img f$

Thus:

 $\displaystyle y$ $\in$ $\displaystyle T$ $\displaystyle \leadsto \ \$ $\displaystyle \exists x \in \Dom f: f \paren x$ $=$ $\displaystyle y$ by hypothesis $\displaystyle \leadsto \ \$ $\displaystyle y$ $\in$ $\displaystyle \Img f$ Definition of Image of Mapping $\displaystyle \leadsto \ \$ $\displaystyle T$ $\subseteq$ $\displaystyle \Img f$ Definition of Subset

Thus by definition of set equality:

$\Img f = T$

and by definition of image of mapping:

$f \sqbrk S = T$

Hence $f$ is a surjection by definition 2.

$\Box$

Definition 2 implies Definition 1

Let $f$ be a mapping which fulfils the condition:

$f \sqbrk S = T$

that is:

$\Img f = T$

Then by definition of set equality:

$T \subseteq \Img f$

Hence:

 $\displaystyle y$ $\in$ $\displaystyle T$ $\displaystyle \implies \ \$ $\displaystyle y$ $\in$ $\displaystyle \Img f$

So by the definition of the image of $f$:

$\exists x \in \Dom f: f \paren x = y$

Hence $f$ is a surjection by definition 1.

$\blacksquare$