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

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Theorem

Let $S$ and $T$ be sets.

Let $\mathcal R_S$ and $\mathcal R_t$ be relations on $S$ and $T$, respectively.

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

$\forall p, q \in S: \left({ p \mathrel{\mathcal R_S} q \iff \phi(p) \mathrel{\mathcal R_T} \phi(q) }\right)$


Then for each $p \in S$:

$\mathcal R_S (p) = \phi^{-1}\left({\mathcal R_T \left({ \phi(p) }\right) }\right)$


Proof

Let $p \in S$.

\(\displaystyle \) \(\) \(\displaystyle x \in \mathcal R_S (p)\) $\quad$ $\quad$
\(\, \displaystyle \iff \, \) \(\displaystyle \) \(\) \(\displaystyle p \mathrel{\mathcal R_S} x\) $\quad$ Definition of the image of $p$ under $\mathcal R_S$ $\quad$
\(\, \displaystyle \iff \, \) \(\displaystyle \) \(\) \(\displaystyle \phi(p) \mathrel{\mathcal R_T} \phi(x)\) $\quad$ Premise $\quad$
\(\, \displaystyle \iff \, \) \(\displaystyle \) \(\) \(\displaystyle \phi(x) \in \mathcal R_T \left({ \phi(p) }\right)\) $\quad$ Definition of the image of $\phi(x)$ under $\mathcal R_T$ $\quad$
\(\, \displaystyle \iff \, \) \(\displaystyle \) \(\) \(\displaystyle x \in \phi^{-1}\left({\mathcal R_T \left({ \phi(p) }\right) }\right)\) $\quad$ Definition of inverse image $\quad$

Thus by the Axiom of Extension:

$\mathcal R_S (p) = \phi^{-1}\left({\mathcal R_T \left({ \phi(p) }\right) }\right)$

$\blacksquare$