Cardinality of Surjection

Theorem
Let $$\left|{S}\right| = n$$.

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

Then $$\left|{T}\right| \le n$$.

Also, $$\left|{T}\right| = n$$ iff $$f$$ is a bijection.

Proof

 * Since $$\left|{S}\right| = n$$, there is a surjection from $$S$$ to $$T$$ iff there is a surjection from $$\mathbb{N}_n$$ to $$T$$ from the definition of cardinality.

So we need consider the case only when $$S = \mathbb{N}_n$$.

For each $$x \in T$$, the set $$f^{-1} \left({\left\{{x}\right\}}\right) \ne \varnothing$$ as $$f$$ is a surjection.

By naturally ordered semigroup: NO 1, $$S = \mathbb{N}_n$$ is well-ordered as it is a subset of $$\mathbb{N}$$, a well-ordered set.

Similarly, as $$f^{-1} \left({\left\{{x}\right\}}\right) \subseteq S$$, $$f^{-1} \left({\left\{{x}\right\}}\right)$$ is also well-ordered so therefore has a minimal element.

So we may define a mapping $$g: T \to S$$:

$$\forall x \in T: g \left({x}\right) =$$ the minimal element of $$f^{-1} \left({\left\{{x}\right\}}\right)$$.

If $$x \in T$$, then $$g \left({x}\right) \in f^{-1} \left({\left\{{x}\right\}}\right)$$, so $$f \left({g \left({x}\right)}\right) = x$$.

Thus $$f \circ g = I_T$$ and by Identity Mapping is a Bijection $$I_T: T \to g \left({T}\right)$$ is a bijection.

By Proper Subset has Fewer Elements, $$\left|{g \left({T}\right)}\right| \le n$$.


 * Let $$\left|{g \left({T}\right)}\right| = m$$. By Set Equivalence an Equivalence Relation, $$\left|{T}\right| = m$$ elements.

Suppose $$m = n$$.

Then by Proper Subset has Fewer Elements, $$g \left({T}\right) = S$$, so $$g: T \to S$$ is a bijection.

Therefore:

$$ $$ $$ $$

So $$f: S \to T$$ is a bijection.