Subset equals Preimage of Image implies Injection
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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 1
Let $f$ be such that:
- $\forall A \in \powerset S: A = \map {\paren {f^\gets \circ f^\to} } A$
In particular, it holds for all subsets of $A$ which are singletons.
Now, consider any $x, y \in A$.
We have:
\(\ds \map f x\) | \(=\) | \(\ds \map f y\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \set {\map f x}\) | \(=\) | \(\ds \set {\map f y}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {f^\to} {\set x}\) | \(=\) | \(\ds \map {f^\to} {\set y}\) | Definition of Direct Image Mapping of Mapping | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \set x\) | \(=\) | \(\ds \map {f^\gets} {\map {f^\to} {\set x} }\) | by hypothesis: $A = \map {\paren {f^\gets \circ f^\to} } A$ | ||||||||||
\(\ds \) | \(=\) | \(\ds \map {f^\gets} {\map {f^\to} {\set y} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \set y\) | by hypothesis: $A = \map {\paren {f^\gets \circ f^\to} } A$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds y\) |
So $f$ is an injection.
$\blacksquare$
Proof 2
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.
$\blacksquare$