Cantor-Bernstein-Schröder Theorem/Proof 5

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

Let $f: S \to T$ and $g: T \to S$ be injections.

Then there exists a bijection $\phi: S \to T$.

Proof
By Injection to Image is Bijection:


 * $g: T \to g \left({T}\right)$ is a bijection.

Thus $T$ is equivalent to $g \left({T}\right)$.

By Composite of Injections is Injection $g \circ f: S \to g \left({T}\right) \subset S$ is also an injection (to a subset of the domain).

Then by Cantor-Bernstein-Schröder Theorem: Lemma:


 * There is a bijection $h: S \to g \left({T}\right)$.

Thus $S$ is equivalent to $g \left({T}\right)$.

We already know that $T$ is equivalent to $g \left({T}\right)$.

Thus by Set Equivalence is Equivalence Relation, $S$ is equivalent to $T$.

By the definition of set equivalence:


 * There is a bijection $\phi: S \to T$.