# 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 $p \in S$.

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

Thus by the Axiom of Extension:

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

$\blacksquare$