Equivalence of Definitions of Injection/Definition 1 iff Definition 4

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

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

$f^{-1} \sqbrk {\set y}$ has more than one element.

That is:
 * $\exists y \in T: x_1, x_2 \in \map {f^{-1} } y, x_1 \ne x_2$

Then we have:
 * $\map f {x_1} = \map f {x_2}$

but:
 * $x_1 \ne x_2$

This contradicts our initial hypothesis that $f$ is an injection by definition 1.

From this contradiction it follows that $f^{-1} \sqbrk {\set y}$ has no more than one element.

That is, $f$ is an injection by definition 4.

Let $f: S \to T$ be an injection by definition 4.

That is, let $\map {f^{-1} } y$ be a singleton for all $y \in T$.

it is not the case that:
 * $\forall x_1, x_2 \in S: \map f {x_1} = \map f {x_2} \implies x_1 = x_2$

Then by definition:
 * $\exists x_1, x_2 \in S, x_1 \ne x_2: \map f {x_1} = \map f {x_2} = y$

By definition of preimage of $y \in T$:
 * $x_1 \in \map {f^{-1} } y, x_2 \in \map {f^{-1} } y$

and so: $\set {x_1, x_2} \subseteq \map {f^{-1} } y$

Thus $\map {f^{-1} } y$ has more than one element for at least one $y \in T$.

This contradicts our initial hypothesis that $f$ is an injection by definition 4.

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

So $f$ is an injection by definition 1.