Difference of Images under Mapping not necessarily equal to Image of Difference

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

The image of the set difference of two subsets of $S$ is not necessarily equal to the set difference of the images.

That is:

Let $S_1$ and $S_2$ be subsets of $S$.

Then it is not always the case that:
 * $f \sqbrk {S_1} \setminus f \sqbrk {S_2} = f \sqbrk {S_1 \setminus S_2}$

where $\setminus$ denotes set difference.

Proof
Note that from Image of Set Difference under Mapping:
 * $f \sqbrk {S_1} \setminus f \sqbrk {S_2} \subseteq f \sqbrk {S_1 \setminus S_2}$

By Proof by Counterexample it is demonstrated that the inclusion does not necessarily apply in the other direction.

Let:
 * $S_1 = \set {x \in \Z: x \le 0}$
 * $S_2 = \set {x \in \Z: x \ge 0}$
 * $f: \Z \to \Z: \forall x \in \Z: f \paren x = x^2$

We have:
 * $S_1 \setminus S_2 = \set {-1, -2, -3, \ldots}$
 * $f \sqbrk {S_1} = \set {0, 1, 4, 9, 16, \ldots} = f \sqbrk {S_2}$

Then from Set Difference with Self is Empty Set:
 * $f \sqbrk {S_1} \setminus f \sqbrk {S_2} = \O$

but:
 * $f \sqbrk {S_1 \setminus S_2} = f \sqbrk {\set {x \in \Z: x > 0} } = \set {1, 4, 9, 16, \ldots}$