Image of Set Difference under Mapping/Corollary 3

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

Let $A \subseteq S$ be a subset of $S$.

Then:
 * $T \setminus f \sqbrk A \subseteq f \sqbrk {T \setminus A}$

where $\setminus$ denotes set difference.

Proof
As $T$ is a surjection, $T = f \sqbrk S$.