Surjection iff Right Inverse/Proof 2

Proof
Take the result Condition for Composite Mapping on Right:

Let $A, B, C$ be sets.

Let $f: B \to A$ and $g: C \to A$ be mappings.

Then:
 * $\Img g \subseteq \Img f$


 * $\exists h: C \to B$ such that $h$ is a mapping and $f \circ h = g$.
 * $\exists h: C \to B$ such that $h$ is a mapping and $f \circ h = g$.

Let $C = A = T$, let $B = S$ and let $g = I_T$.

Then the above translates into:


 * $\Img {I_T} \subseteq \Img f$


 * $\exists g: T \to S$ such that $g$ is a mapping and $f \circ g = I_T$.
 * $\exists g: T \to S$ such that $g$ is a mapping and $f \circ g = I_T$.

But we know that:
 * $\Img f \subseteq T = \Img {I_T}$

So by definition of set equality, the result follows.