Cardinality of Image of Mapping not greater than Cardinality of Domain

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

Let $f: S \to T$ be a mapping.

Let $\left|{S}\right|$ denote the cardinal number of $S$.

Let $S \sim \left|{S}\right|$.

Then:


 * $\left\vert{\operatorname{Im} \left({f}\right)}\right\vert \le \left\vert{S}\right\vert$

Proof
By Restriction of Mapping to Image is Surjection, the mapping:
 * $f: S \to \operatorname{Im} \left({f}\right)$

is a surjection.

Let $h$ be a mapping such that:
 * $h: \left|{S}\right| \to S$

is a bijection.

By Composite of Surjections is Surjection:
 * $f \circ h : \left|{ S }\right| \to \operatorname{Im}\left({f}\right)$

is a surjection.

Construct a set $R$ such that:


 * $R = \left\{ x \in \left|{ S }\right| : \forall y \in x: f\left({h\left({x}\right)}\right) \ne f\left({h\left({y}\right)}\right) \right\}$

It follows that $R \subseteq \left|{ S }\right|$.

By Subset of Ordinal implies Cardinal Inequality, $\left|{ R }\right| \le \left|{ S }\right|$.

Suppose $x \in R \land y \in R \land f \left({ h \left({ x }\right)}\right) = f\left({h\left({y}\right)}\right)$.

Then, $x$ and $y$ are ordinals and $x < y \lor x = y \lor y < x$ by Ordinal Membership is Trichotomy.

If $x < y$, then $f\left({h\left({x}\right)}\right) \ne f\left({h\left({y}\right)}\right)$ by the definition of $R$.

Similarly, $y < x$ implies that $f\left({h\left({x}\right)}\right) \ne f\left({h\left({y}\right)}\right)$.

Therefore, $x = y$.

It follows that the restriction $f \circ h \restriction_R : R \to \operatorname{Im}\left({f}\right)$ is an injection.

Finally, for all $x \in \operatorname{Im}\left({f}\right)$, there is some $y \in \left|{ S }\right|$ such that $f\left({h\left({y}\right)}\right) = x$ by the definition of surjection.

But since this is true for some $y$, the set:


 * $\left\{ y \in \left|{ S }\right| : f\left({h\left({y}\right)}\right) = x \right\}$ has a minimal element.

For this minimal element $y$, it follows that:


 * $\forall z \in y: f\left({h\left({z}\right)}\right) \ne f\left({h\left({y}\right)}\right)$, since if it were equal, this would contradict the fact that $y$ is a $\in$-minimal element.

It follows that the restriction $f \circ h \restriction_R : R \to \operatorname{Im}\left({f}\right)$ is a bijection.

By the definition of set equivalence, $R \sim \operatorname{Im}\left({f}\right)$.

So: