Inverse of Composite Relation

Theorem
Let $$\mathcal R_2 \circ \mathcal R_1 \subseteq S_1 \times S_3$$ be the composite of the two relations $$\mathcal R_1 \subseteq S_1 \times S_2$$ and $$\mathcal R_2 \subseteq S_2 \times S_3$$.

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

Proof
Let $$\mathcal R_1 \subseteq S_1 \times S_2$$ and $$\mathcal R_2 \subseteq S_2 \times S_3$$ be relations.

We assume that:
 * $$\operatorname{Dom} \left({\mathcal R_2}\right) = \operatorname{Cdm} \left({\mathcal R_1}\right)$$

where $$\operatorname{Dom}$$ denotes domain and $$\operatorname{Cdm}$$ denotes codomain.

This is necessary for $$\mathcal R_2 \circ \mathcal R_1$$ to exist.

From the definition of an inverse relation, we have:


 * $$\operatorname{Dom} \left({\mathcal R_2}\right) = \operatorname{Cdm} \left({\mathcal R_2^{-1}}\right)$$
 * $$\operatorname{Cdm} \left({\mathcal R_1}\right) = \operatorname{Dom} \left({\mathcal R_1^{-1}}\right)$$

So we confirm that $$\mathcal R_1^{-1} \circ \mathcal R_2^{-1}$$ is defined.

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