# Preimage of Image of Subset under Injection equals Subset

## Theorem

Let $f: S \to T$ be an injection.

Then:

$\forall A \subseteq S: A = \paren {f^{-1} \circ f} \sqbrk A$

where:

$f \sqbrk A$ denotes the image of $A$ under $f$
$f^{-1}$ denotes the inverse of $f$
$f^{-1} \circ f$ denotes composition of $f^{-1}$ and $f$.

## Proof

Let $f$ be an injection.

From Subset of Domain is Subset of Preimage of Image, we have that:

$\forall A \subseteq S: A \subseteq \paren {f^{-1} \circ f} \sqbrk A$

by dint of $f$ being a relation.

So what we need to do is show that:

$\forall A \subseteq S: \paren {f^{-1} \circ f} \sqbrk A \subseteq A$

Take any $A \subseteq S$.

Let $x \in A$.

We have:

 $\displaystyle x$ $\in$ $\displaystyle \paren {f^{-1} \circ f} \sqbrk A$ $\displaystyle \leadsto \ \$ $\displaystyle x$ $\in$ $\displaystyle f^{-1} \sqbrk {f \sqbrk A}$ Definition of Composition of Mappings $\displaystyle \leadsto \ \$ $\displaystyle \map f x$ $\in$ $\displaystyle f \sqbrk A$ Definition of Inverse of Mapping $\displaystyle \leadsto \ \$ $\displaystyle \exists y \in A: \ \$ $\displaystyle \map f x$ $=$ $\displaystyle \map f y$ Definition of Image of Subset under Mapping $\displaystyle \leadsto \ \$ $\displaystyle x$ $=$ $\displaystyle y$ Definition of Injection $\displaystyle \leadsto \ \$ $\displaystyle x$ $\in$ $\displaystyle A$ as $y \in A$

Thus we see that:

$\paren {f^{-1} \circ f} \sqbrk A \subseteq A$

and hence the result:

$\forall A \subseteq S: A = \paren {f^{-1} \circ f} \sqbrk A$

$\blacksquare$