Subset equals Preimage of Image implies Injection/Proof 2

Theorem

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

Let $f^\to: \powerset S \to \powerset T$ be the direct image mapping of $f$.

Similarly, let $f^\gets: \powerset T \to \powerset S$ be the inverse image mapping of $f$.

Let:

$\forall A \in \powerset S: A = \map {\paren {f^\gets \circ f^\to} } A$

Then $f$ is an injection.

Proof

Suppose that $f$ is not an injection.

Then two elements of $S$ map to the same one element of $T$.

That is:

$\exists a_1, a_2 \in S, b \in T: \map f {a_1} = \map f {a_2} = b$

Let $A = \set {a_1}$.

Then:

\(\ds \map {f^\to} A\)

\(=\)

\(\ds \set b\)

\(\ds \leadsto \ \ \)

\(\ds \map {f^\gets} {\map {f^\to} A}\)

\(\supseteq\)

\(\ds \set {a_1, a_2}\)

Definition of Preimage of Subset under Mapping

\(\ds \leadsto \ \ \)

\(\ds \map {f^\gets} {\map {f^\to} A}\)

\(\supsetneqq\)

\(\ds A\)

as $A = \set {a_1}$

So by the Rule of Transposition:

$\forall A \in \powerset S: A = \map {\paren {f^\gets \circ f^\to} } A$

implies that $f$ is an injection.

$\blacksquare$

Sources

• 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $2 \ \text{(a})$