Equivalence of Definitions of Mapping

Proof
In the following, let $f \subseteq S \times T$ be a relation on $S \times T$.

$(1)$ $(4)$
$f$ is by definition left-total :
 * $\forall s \in S: \exists t \in T: \tuple {s, t} \in f$

which means exactly the same thing as:
 * each element of $S$ is associated with with at least one element of $T$.

$f$ is by definition many-to-one :
 * $\forall x \in \Dom f: \tuple {x, y_1} \in f \land \tuple {x, y_2} \in f \implies y_1 = y_2$

which means exactly the same thing as:
 * each element of $S$ is associated with with at most one element of $T$.

Thus:
 * $f$ is left-total and many-to-one


 * each element of $S$ is associated by $f$ with with exactly one one element of $T$.
 * each element of $S$ is associated by $f$ with with exactly one one element of $T$.

That is:
 * $f$ is a mapping by definition 1


 * $f$ is a mapping by definition 4.
 * $f$ is a mapping by definition 4.

$(2)$ $(4)$
$f$ is by definition left-total :
 * $\forall s \in S: \exists t \in T: \tuple {s, t} \in \mathcal R$

and:

$f$ is by definition many-to-one :
 * $\forall x \in \Dom f: \forall y_1, y_2 \in \Cdm f: \tuple {x, y_1} \in f \land \tuple {x, y_2} \in f \implies y_1 = y_2$

which means exactly the same thing as:
 * $\forall x \in S: \forall y_1, y_2 \in T: \tuple {x, y_1} \in f \land \tuple {x, y_2} \in f \implies y_1 = y_2$

That is:
 * $f$ is a mapping by definition 2


 * $f$ is a mapping by definition 4.
 * $f$ is a mapping by definition 4.

$(2)$ $(3)$
Both definition 2 and definition 3 carry the stipulation that:


 * $\forall x \in S: \exists y \in T: \tuple {x, y} \in f$

Then we have that definition 2 carries the stipulation that


 * $\forall x \in S: \forall y_1, y_2 \in T: \tuple {x, y_1} \in f \land \tuple {x, y_2} \in f \implies y_1 = y_2$

By the Rule of Transposition, that means the same as:


 * $\forall x \in S: \forall y_1, y_2 \in T: \neg \paren {y_1 = y_2} \implies \neg \paren {\tuple {x, y_1} \in f \land \tuple {x, y_2} \in f}$

which in turn is the same as:


 * $\forall \tuple {x_1, y_1}, \tuple {x_2, y_2} \in f: y_1 \ne y_2 \implies x_1 \ne x_2$

which is the other stipulation of definition 3.

That is:
 * $f$ is a mapping by definition 2


 * $f$ is a mapping by definition 3.
 * $f$ is a mapping by definition 3.