Subset of Domain is Subset of Preimage of Image

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Theorem

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


Then:

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

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 $f$.


This can be expressed in the language and notation of direct image mappings and inverse image mappings as:

$\forall A \in \powerset S: A \subseteq \map {\paren {f^\gets \circ f^\to} } A$


Equality does Not Necessarily Hold

It is not necessarily the case that:

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


Proof

As a mapping is by definition a left-total relation.

Therefore Preimage of Image under Left-Total Relation is Superset applies:

$A \subseteq S \implies A \subseteq \paren {\RR^{-1} \circ \RR} \sqbrk A$

where $\RR$ is a relation.

Hence:

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

$\blacksquare$


Sources

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