Composition of Direct Image Mappings of Relations

Theorem
Let $A, B, C$ be non-empty sets.

Let $\mathcal R_1 \subseteq A \times B, \mathcal R_2 \subseteq B \times C$ be relations.

Let:
 * ${\mathcal R_1}^\to: \mathcal P \left({A}\right) \to \mathcal P \left({B}\right)$

and
 * ${\mathcal R_2}^\to: \mathcal P \left({B}\right) \to \mathcal P \left({C}\right)$

be the mappings induced by $\mathcal R_1$ and $\mathcal R_2$.

Let:
 * ${\mathcal R_1}^\gets: \mathcal P \left({B}\right) \to \mathcal P \left({A}\right)$

and
 * ${\mathcal R_2}^\gets: \mathcal P \left({C}\right) \to \mathcal P \left({B}\right)$

be the mappings induced by their inverse relations $\mathcal R_1$ and $\mathcal R_2$.

Then:
 * $\left({\mathcal R_2 \circ \mathcal R_1}\right)^\to = {\mathcal R_2}^\to \circ {\mathcal R_1}^\to$
 * $\left({\mathcal R_2 \circ \mathcal R_1}\right)^\gets = {\mathcal R_1}^\gets \circ {\mathcal R_2}^\gets$

Proof
Let $S \subseteq A, S \ne \varnothing$.

Then:

Now we treat the case where $S = \varnothing$:

Now consider $T \subseteq C$.

We have:


 * $\left({{\mathcal R_1}^\gets \circ {\mathcal R_2}^\gets}\right) \left({T}\right) = \begin{cases}

\varnothing & : \operatorname{Im} \left({R_1}\right) \cap {\mathcal R_2}^\gets \left({T}\right) = \varnothing \\ \left\{{x \in A: \mathcal R_1 \left({x}\right) \in {\mathcal R_2}^\gets \left({T}\right)}\right\} & : \text {otherwise} \end{cases}$


 * $\left({\mathcal R_2 \circ \mathcal R_1}\right)^\gets \left({T}\right) = \begin{cases}

\varnothing & : \operatorname{Im} \left({R_2 \circ R_1}\right) \cap T = \varnothing \\ \left\{{x \in A: \mathcal R_2 \left({\mathcal R_1 \left({x}\right)}\right) \in T}\right\} & : \text {otherwise} \end{cases}$

We need to show that:
 * $\operatorname{Im} \left({R_1}\right) \cap {\mathcal R_2}^\gets \left({T}\right) = \varnothing \iff \operatorname{Im} \left({R_2 \circ R_1}\right) \cap T = \varnothing$

So: