Intersection of Injective Image with Relative Complement

Theorem

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

Then $f$ is an injection if and only if:

$\forall A \subseteq S: f \sqbrk A \cap f \sqbrk {\relcomp S A} = \O$

Proof

From Intersection with Relative Complement is Empty:

$A \cap \relcomp S A = \O$

From Image of Intersection under Injection:

$\forall A, B \subseteq S: f \sqbrk {A \cap B} = f \sqbrk A \cap f \sqbrk B$

if and only if $f$ is an injection.

Hence the result.

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $4$: Mappings: Exercise $8$