# Subset equals Preimage of Image implies Injection

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## Contents

## 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:

\(\displaystyle \map f x\) | \(=\) | \(\displaystyle \map f y\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \set {\map f x}\) | \(=\) | \(\displaystyle \set {\map f y}\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map {f^\to} {\set x}\) | \(=\) | \(\displaystyle \map {f^\to} {\set y}\) | Definition of Direct Image Mapping of Mapping | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \set x\) | \(=\) | \(\displaystyle \map {f^\gets} {\map {f^\to} {\set x} }\) | by hypothesis: $A = \map {\paren {f^\gets \circ f^\to} } A$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map {f^\gets} {\map {f^\to} {\set y} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \set y\) | by hypothesis: $A = \map {\paren {f^\gets \circ f^\to} } A$ | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x\) | \(=\) | \(\displaystyle y\) |

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:

\(\displaystyle \map {f^\to} A\) | \(=\) | \(\displaystyle \set b\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map {f^\gets} {\map {f^\to} A}\) | \(\supseteq\) | \(\displaystyle \set {a_1, a_2}\) | Definition of Preimage of Subset under Mapping | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map {f^\gets} {\map {f^\to} A}\) | \(\supsetneqq\) | \(\displaystyle A\) | 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.

$\blacksquare$