Inverses of Right-Total and Left-Total Relations

Theorem
Let $\mathcal R \subseteq S \times T$ be a relation on $S \times T$.

Let $\mathcal R^{-1} \subseteq T \times S$ be the inverse of $\mathcal R$.

Then $\mathcal R$ is right-total iff $\mathcal R^{-1}$ is left-total.

Similarly, $\mathcal R$ is left-total iff $\mathcal R^{-1}$ is right-total.

Proof
Let $\mathcal R$ be right-total.

Then by definition:
 * $\forall t \in T: \exists s \in S: \left({s, t}\right) \in \mathcal R$

By definition of the inverse of $\mathcal R$, it follows that:
 * $\forall t \in T: \exists s \in S: \left({t, s}\right) \in \mathcal R^{-1}$

So by definition $\mathcal R^{-1}$ is left-total.

The argument reverses: let $\mathcal R^{-1}$ is left-total.

Then by definition:
 * $\forall t \in T: \exists s \in S: \left({t, s}\right) \in \mathcal R^{-1}$

and so:
 * $\forall t \in T: \exists s \in S: \left({s, t}\right) \in \mathcal R$

As the inverse of $\mathcal R^{-1}$ is $\mathcal R$, the other statement is even more obvious.