Inverse of Many-to-One Relation is One-to-Many

Theorem
The inverse of a many-to-one relation is a one-to-many relation, and vice versa.

Proof
From the definition, a many-to-one relation $\mathcal R \subseteq S \times T$ is such that:


 * $\mathcal R\subseteq S \times T: \forall x \in \operatorname{Dom} \left({\mathcal R}\right): \left({x, y_1}\right) \in \mathcal R \land \left({x, y_2}\right) \in \mathcal R \implies y_1 = y_2$

Also from the definition, a one-to-many relation $\mathcal R \subseteq S \times T$ is such that:


 * $\mathcal R \subseteq S \times T: \forall y \in \operatorname{Im} \left({\mathcal R}\right): \left({x_1, y}\right) \in \mathcal R \land \left({x_2, y}\right) \in \mathcal R \implies x_1 = x_2$

The inverse of a relation $\mathcal R$ is defined as:


 * $\mathcal R^{-1} = \left\{{\left({t, s}\right): \left({s, t}\right) \in \mathcal R}\right\}$

Also from the definition of inverse relation, we have that:
 * $\operatorname{Dom} \left({\mathcal R}\right) = \operatorname{Rng} \left({\mathcal R^{-1}}\right)$
 * $\operatorname{Rng} \left({\mathcal R}\right) = \operatorname{Dom} \left({\mathcal R^{-1}}\right)$

So, let $\mathcal R$ be a many-to-one relation.

Putting the above all together, we have:


 * $\mathcal R^{-1} \subseteq T \times S: \forall x \in \operatorname{Im} \left({\mathcal R^{-1}}\right): \left({y_1, x}\right) \in \mathcal R^{-1} \land \left({y_2, x}\right) \in \mathcal R^{-1} \implies y_1 = y_2$.

... and it can be seen that $\mathcal R^{-1}$ is one-to-many.

Similarly, let $\mathcal R$ be a many-to-one relation. This gives us:


 * $\mathcal R^{-1} \subseteq T \times S: \forall y \in \operatorname{Dom} \left({\mathcal R^{-1}}\right): \left({y, x_1}\right) \in \mathcal R^{-1} \land \left({y, x_2}\right) \in \mathcal R^{-1} \implies x_1 = x_2$.

... and it can be seen that $\mathcal R^{-1}$ is many-to-one.