Equivalence of Definitions of Surjection

Definition 1 implies Definition 2
Let $f$ be a mapping which fulfils the condition:
 * $\forall y \in T: \exists x \in \Dom f: f \paren x = y$

From Image is Subset of Codomain:
 * $\Img f \subseteq T$

It remains to be proved that:
 * $T \subseteq \Img f$

Thus:

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.

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:

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.