# Image of Set Difference under Injection

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## 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:

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

if and only if $f$ 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:

- $\RR \sqbrk {S_1} \setminus \RR \sqbrk {S_2} = \RR \sqbrk {S_1 \setminus S_2}$

if and only if $\RR$ 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 if and only if $f$ is also an injection.

It follows that:

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

if and only if $f$ is an injection.

$\blacksquare$

## Sources

- 1971: Robert H. Kasriel:
*Undergraduate Topology*... (previous) ... (next): $\S 1.12$: Set Inclusions for Image and Inverse Image Sets: Exercise $4 \ \text{(a)}$ - 2000: James R. Munkres:
*Topology*(2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.2 \ \text{(h)}$ - 2005: René L. Schilling:
*Measures, Integrals and Martingales*... (previous) ... (next): $2.2$

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- 1960: Paul R. Halmos:
*Naive Set Theory*... (previous) ... (next): $\S 10$: Inverses and Composites: Exercise $\text{(iii)}$