Inverse Image under Embedding of Image under Relation of Image of Point

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Theorem

Let $S$ and $T$ be sets.

Let $\RR_S$ and $\RR_t$ be relations on $S$ and $T$, respectively.

Let $\phi: S \to T$ be a mapping with the property that:

$\forall p, q \in S: \paren {p \mathrel {\RR_S} q \iff \map \phi p \mathrel {\RR_T} \map \phi q}$

Then for each $p \in S$:

$\map {\RR_S} p = \phi^{-1} \sqbrk {\map {\RR_T} {\map \phi p} }$

Let $p \in S$.

$\ds x$

$\in$

$\ds \map {\RR_S} p$

$\ds \leadstoandfrom \ \ $

$\ds p$

$\mathrel {\RR_S}$

$\ds x$

Definition of Image of $p$ under $\RR_S$

$\ds \leadstoandfrom \ \ $

$\ds \map \phi p$

$\mathrel {\RR_T}$

$\ds \map \phi x$

$\ds \leadstoandfrom \ \ $

$\ds \map \phi x$

$\in$

$\ds \map {\RR_T} {\map \phi p}$

Definition of the image of $\map \phi x$ under $\RR_T$

$\ds \leadstoandfrom \ \ $

$\ds x$

$\in$

$\ds \phi^{-1} \sqbrk {\map {\RR_T} {\map \phi p} }$

Definition of inverse image

Thus by definition of set equality:

$\map {\RR_S} p = \phi^{-1} \sqbrk {\map {\RR_T} {\map \phi p} }$

$\blacksquare$