Image of Set Difference under Mapping/Corollary 3

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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:

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

$\blacksquare$


Sources