Between Two Sets Exists Injection or Surjection

Theorem
Let $S$ and $T$ be sets.

Then either or both of the following cases hold:
 * $(1):$ There exists a mapping $f: S \to T$ such that $f$ is an injection
 * $(2):$ There exists a mapping $f: S \to T$ such that $f$ is a surjection.

Proof
From Zermelo's Theorem, exactly one of the following cases holds:
 * $(1): \quad S \le T$, that is, $T$ dominates $S$
 * $(2): \quad S \equiv T$, that is, $T$ is equivalent to $S$
 * $(3): \quad S \ge T$, that is, $S$ dominates $T$.

Suppose $(1)$ holds, that is, $T$ dominates $S$.

By definition, there exists an injection from $S$ to $T$.

Hence in this case the result holds.

Suppose $(2)$ holds, that is, $T$ is equivalent to $S$.

By definition, there then exists a bijection from $S$ to $T$.

A bijection is, by definition, a mapping which is both an injection and a surjection.

Hence in this case the result holds.

Suppose $(3)$ holds, that is, $S$ dominates $T$.

By definition, there exists an injection from $S$ to $T$.

By Injection iff Left Inverse, $f$ has a left inverse $g: T \to S$ such that $g \circ f = I_S$.

By Left Inverse Mapping is Surjection, $g$ is a surjection.

Hence in this case the result holds.

All cases have been shown to lead to the same conclusion, that is, that there exists either an injection or a surjection or both between $S$ and $T$.

Hence the result, by Proof by Cases.