# 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:

$T \setminus f \sqbrk A \subseteq f \sqbrk {S \setminus A}$

where $\setminus$ denotes set difference.

## Proof

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

Thus Image of Set Difference under Mapping can be applied:

$f \sqbrk S \setminus f \sqbrk A \subseteq f \sqbrk {S \setminus A}$

$\blacksquare$