# Subset equals Preimage of Image implies Injection

## 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.

$\blacksquare$