Equivalence of Definitions of Injection/Definition 1 iff Definition 4

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

Thus:
 * $\forall x_1, x_2 \in S: f \left({x_1}\right) = f \left({x_2}\right) \implies x_1 = x_2$

Suppose that $f^{-1} \left[{\left\{{t}\right\}}\right]$ has more than one element.

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

Then we have:
 * $f \left({x_1}\right) = f \left({x_2}\right)$

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 $f^{-1} \left[{\left\{{y}\right\}}\right]$ 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 $f^{-1} \left({y}\right)$ be a singleton for all $y \in T$.

Suppose it is not the case that:
 * $\forall x_1, x_2 \in S: f \left({x_1}\right) = f \left({x_2}\right) \implies x_1 = x_2$

Then by definition:
 * $\exists x_1, x_2 \in S, x_1 \ne x_2: f \left({x_1}\right) = f \left({x_2}\right) = y$

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

and so: $\left\{{x_1, x_2}\right\} \subseteq f^{-1} \left({y}\right)$

Thus $f^{-1} \left({y}\right)$ 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:
 * $\forall x_1, x_2 \in S: f \left({x_1}\right) = f \left({x_2}\right) \implies x_1 = x_2$

So $f$ is an injection by definition 1.