Subset of Domain is Subset of Preimage of Image/Equality does Not Necessarily Hold

Theorem
Let $f: S \to T$ be a mapping.

From Subset of Domain is Subset of Preimage of Image:
 * $A \subseteq S \implies A \subseteq \paren {f^{-1} \circ f} \sqbrk A$

where:
 * $f \sqbrk A$ denotes the image of $A$ under $f$
 * $f^{-1} \sqbrk A$ denotes the preimage of $A$ under $f$
 * $f^{-1} \circ f$ denotes composition of $f^{-1}$ and $f$.

It is not necessarily the case that:
 * $A \subseteq S \implies A = \paren {f^{-1} \circ f} \sqbrk A$

Proof
Proof by Counterexample:

Let:
 * $S = \set {0, 1}$
 * $T = \set 2$

Let $f: S \to T$ be defined as:
 * $\map f 0 = 2$
 * $\map f 1 = 2$

Let $A \subseteq S$ be defined as:
 * $A = \set 0$

Then we have:
 * $f \sqbrk A = \set 2$

but:
 * $f^{-1} \circ f \sqbrk A = f^{-1} \sqbrk 2 = \set {0, 1}$

That is:
 * $A \subseteq \paren {f^{-1} \circ f} \sqbrk A$

but it is not the case that:
 * $A \paren {f^{-1} \circ f} \sqbrk A$