# Equivalence of Definitions of Surjection

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## 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:

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

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

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.

$\blacksquare$