# Inverse Image under Embedding of Image under Relation of Image of Point It has been suggested that this article or section be renamed. One may discuss this suggestion on the talk page.

## 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:

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

$\blacksquare$