Image of Set Difference under Injection

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

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

Let $S_1 \setminus S_2$ denote the set difference between $S_1$ and $S_2$.

Then:
 * $\forall S_1, S_2 \subseteq S: f \sqbrk {S_1} \setminus f \sqbrk {S_2} = f \sqbrk {S_1 \setminus S_2}$

$f$ is an injection.

Proof
An injection is a type of one-to-one relation, and therefore also a one-to-many relation.

Therefore One-to-Many Image of Set Difference applies:
 * $\RR \sqbrk {S_1} \setminus \RR \sqbrk {S_2} = \RR \sqbrk {S_1 \setminus S_2}$

$\RR$ is one-to-many.

We have that $f$ is a mapping and therefore a many-to-one relation.

So $f$ is a one-to-many relation $f$ is also an injection.

It follows that:
 * $\forall S_1, S_2 \subseteq S: f \sqbrk {S_1} \setminus f \sqbrk {S_2} = f \sqbrk {S_1 \setminus S_2}$

$f$ is an injection.