Element in Preimage of Image under Mapping

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $f^{-1} \sqbrk {\map f x}$ denote the preimage of $\map f x$ under $f$.


Then:

$\forall x \in S: x \in f^{-1} \sqbrk {\map f x}$


Proof

A mapping is by definition a left-total relation.

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

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

Thus:

$\set x \subseteq S \implies \set x \subseteq f^{-1} \sqbrk {f \sqbrk A}$

Hence the result.

$\blacksquare$


Sources