Equivalence of Definitions of Injection/Definition 1 iff Definition 3

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

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

First we note that:
 * $t \in \operatorname{Im} \left({f}\right) \implies \exists x \in \operatorname{Dom} \left({f}\right): f \left({x}\right) = t$

thus fulfilling the condition for $f^{-1} {\restriction_{\operatorname{Im} \left({f}\right)} }$ to be left-total.

Now let:
 * $t \in \operatorname{Im} \left({f}\right): \left({t, y}\right), \left({t, z}\right) \in f^{-1}$

Thus:

So by the definition of mapping, $f^{-1} {\restriction_{\operatorname{Im} \left({f}\right)} }$ is a mapping.

So $f$ is an injection by definition 3.

Let $f: S \to T$ be an injection by definition 3.

Then:
 * $f^{-1} {\restriction_{\operatorname{Im} \left({f}\right)} }: \operatorname{Im} \left({f}\right) \to \operatorname{Dom} \left({f}\right)$ is a mapping

where $f^{-1} {\restriction_{\operatorname{Im} \left({f}\right)} }$ is the restriction of the inverse of $f$ to the image set of $f$.

We need to show that:
 * $\forall x, z \in \operatorname{Dom} \left({f}\right): f \left({x}\right) = f \left({z}\right) \implies x = z$

So, pick any $x, z \in \operatorname{Dom} \left({f}\right)$ such that:
 * $f \left({x}\right) = f \left({z}\right)$

Then:

So $f$ is an injection by definition 1.