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:

Thus we see that:
 * $\paren {f^{-1} \circ f} \sqbrk A \subseteq A$

and hence the result:
 * $\forall A \subseteq S: A = \paren {f^{-1} \circ f} \sqbrk A$

Also see

 * Subset equals Preimage of Image implies Injection
 * Subset equals Preimage of Image iff Mapping is Injection