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:


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

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


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

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:


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

... 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:


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

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