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.