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

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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:

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

$\blacksquare$


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