Equivalence of Definitions of Injection/Definition 1 iff Definition 2

Proof
Let $f: S \to T$ be an injection by definition 1.

Thus:
 * $(1): \quad \forall x_1, x_2 \in S: \map f {x_1} = \map f {x_2} \implies x_1 = x_2$

Let $y \in T: y = \map f {x_1} = \map f {x_2}$.

Thus $(1)$ can be rewritten in the language of relations as:
 * $\forall x_1, x_2 \in S: \tuple {x_1, y} \in f \land \tuple {x_2, y} \in f \implies x_1 = x_2$

So, by definition, $f$ is a one-to-many relation.

By definition, $f$ is also mapping.

So, by definition, $f$ is a relation which is both many-to-one and left-total.

A relation which is both many-to-one and one-to-many is by definition a one-to-one relation.

Thus $f$ is a relation which is one-to-one and left-total.

So $f$ is an injection by definition 2.

Let $f: S \to T$ be an injection by definition 2.

That is, $f$ is a relation which is one-to-one and left-total.

A one-to-one relation is a relation which is both many-to-one and one-to-many.

So $f$ is a relation which is both many-to-one and left-total.

Thus $f$ is a mapping which is one-to-many:
 * $\forall x_1, x_2 \in S: \tuple {x_1, y} \in f \land \tuple {x_2, y} \in f \implies x_1 = x_2$

Setting $y = \map f {x_1} = \map f {x_2}$:
 * $\forall x_1, x_2 \in S: \map f {x_1} = \map f {x_2} \implies x_1 = x_2$

So $f$ is an injection by definition 1.