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 $$f_{\mathcal R_1}: \mathcal P \left({A}\right) \to \mathcal P \left({B}\right)$$ and $$f_{\mathcal R_2}: \mathcal P \left({B}\right) \to \mathcal P \left({C}\right)$$ be the mappings induced by $$\mathcal R_1$$ and $$\mathcal R_2$$.

Then:
 * $$f_{\left({\mathcal R_2 \circ \mathcal R_1}\right)} = f_{\mathcal R_2} \circ f_{\mathcal R_1}$$
 * $$f_{\left({\mathcal R_2 \circ \mathcal R_1}\right)^{-1}} = f_{\mathcal R_1^{-1}} \circ f_{\mathcal R_2^{-1}}$$

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

Then:

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

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Now consider $$T \subseteq C$$.

We have:


 * $$\left({f_{\mathcal R_1^{-1}} \circ f_{\mathcal R_2^{-1}}}\right) \left({T}\right) = \begin{cases}

\varnothing & : \operatorname{Im} \left({R_1}\right) \cap f_{\mathcal R_2^{-1}} \left({T}\right) = \varnothing \\ \left\{{x \in A: \mathcal R_1 \left({x}\right) \in f_{\mathcal R_2^{-1}}\left({T}\right)}\right\} & : \text {otherwise} \end{cases}$$


 * $$f_{\left({\mathcal R_2 \circ \mathcal R_1}\right)^{-1}} \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:
 * $$\operatorname{Im} \left({R_1}\right) \cap f_{\mathcal R_2^{-1}} \left({T}\right) = \varnothing \iff \operatorname{Im} \left({R_2 \circ R_1}\right) \cap T = \varnothing$$

So:

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