Inverse of Injective and Surjective Mapping is Mapping/Proof 2

Proof
Let $f: S \to T$ be a mapping such that:
 * $(1): \quad f$ is an injection
 * $(2): \quad f$ is a surjection.

Let $t \in T$.

Then as $f$ is a surjection:


 * $\exists s \in S: t = f \paren s$

As $f$ is an injection, there is only one $s \in S$ such that $t = f \paren s$.

Define $g \paren t = s$.

As $t \in T$ is arbitrary, it follows that:


 * $\forall t \in T: \exists s \in S: g \paren t = s$

such that $s$ is unique for a given $t$.

That is, $g: T \to S$ is a mapping.

By the definition of $g$:


 * $(1): \quad \forall t \in T: f \paren {g \paren t} = t$

Let $s \in S$.

Let:
 * $(2): \quad s' = g \paren {f \paren s}$.

Then:

Thus $f: S \to T$ and $g: T \to S$ are inverse mappings of each other.