Condition for Composite Relation with Inverse to be Identity

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

Let $$\mathcal R \circ \mathcal R^{-1}$$ be the composite of $$\mathcal R$$ with its inverse.

Let $$I_T$$ be the identity mapping on $$T$$.

Then:
 * $$\mathcal R \circ \mathcal R^{-1} = I_T$$

iff:
 * $$\mathcal R$$ is functional;
 * $$\mathcal R$$ is right-total.

Example

 * CompositeWithInverseIdentity.png

Note in the above that:


 * $$\mathcal R$$ is functional;
 * $$\mathcal R$$ is right-total;
 * $$\mathcal R \circ \mathcal R^{-1} = I_T$$.

Note, however, that $$\mathcal R^{-1}$$ is neither functional nor right-total, and does not need to be for $$\mathcal R \circ \mathcal R^{-1} = I_T$$.

Necessary Condition
Let:
 * $$\mathcal R$$ be functional;
 * $$\mathcal R$$ be right-total.

$$\operatorname{Im} \left({\mathcal R}\right) = T$$.

Let $$\left({t_1, t_2}\right) \in \mathcal R \circ \mathcal R^{-1}$$.

The composite of $\mathcal R^{-1}$ and $\mathcal R$ is defined as:


 * $$\mathcal R \circ \mathcal R^{-1} = \left\{{\left({t_1, t_2}\right) \in T \times T: \exists s \in S: \left({t_1, s}\right) \in \mathcal R^{-1} \and \left({s, t_2}\right) \in \mathcal R}\right\}$$

By definition of inverse:
 * $$\mathcal R \circ \mathcal R^{-1} = \left\{{\left({t_1, t_2}\right) \in T \times T: \exists s \in S: \left({s, t_1}\right) \in \mathcal R \and \left({s, t_2}\right) \in \mathcal R}\right\}$$

But $$\mathcal R$$ is functional, and so $$t_1 = t_2$$.

So:
 * $$\forall \left({t_1, t_2}\right) \in \mathcal R \circ \mathcal R^{-1}: t_1 = t_2$$

and so:
 * $$\mathcal R \circ \mathcal R^{-1} \subseteq I_T$$

Now let $$t \in T$$.

By definition of identity mapping on $$T$$, we have that $$\left({t, t}\right) \in I_T$$.

As $$\mathcal R$$ is right-total, we have that $$\operatorname{Im} \left({\mathcal R}\right) = T$$, and so:
 * $$\exists s \in S: \left({s, t}\right) \in \mathcal R$$

and so:
 * $$\exists s \in S: \left({t, s}\right) \in \mathcal R^{-1}$$

Hence by definition of the composite of $\mathcal R^{-1}$ and $\mathcal R$:
 * $$\left({t, t}\right) \in \mathcal R \circ \mathcal R^{-1}$$

So:
 * $$\mathcal R \circ \mathcal R^{-1} \supseteq I_T$$

and so:
 * $$\mathcal R \circ \mathcal R^{-1} = I_T$$

Sufficient Condition
Let $$\mathcal R \circ \mathcal R^{-1} = I_T$$.


 * Suppose $$\exists t \in T: t \notin \operatorname{Im} \left({\mathcal R}\right)$$.

Then $$t \notin \operatorname{Im} \left({\mathcal R \circ \mathcal R^{-1}}\right)$$.

But $$t \in \operatorname{Im} \left({I_T}\right)$$ by definition of the identity mapping on $$T$$.

Hence $$\mathcal R \circ \mathcal R^{-1} \ne I_T$$.

From this contradiction we deduce that:
 * $$\mathcal R \circ \mathcal R^{-1} = I_T \implies T \backslash \operatorname{Im} \left({\mathcal R}\right) = \varnothing$$

where $$T \backslash \operatorname{Im} \left({\mathcal R}\right)$$ denotes set difference.

So from Set Difference with Superset is Empty Set‎:
 * $$T \subseteq \operatorname{Im} \left({\mathcal R}\right)$$

But from Image is Subset of Codomain we have:
 * $$T \supseteq \operatorname{Im} \left({\mathcal R}\right)$$

and so:
 * $$\operatorname{Im} \left({\mathcal R}\right) = T$$

which means $$\mathcal R$$ is right-total.


 * Suppose $$\mathcal R$$ is not functional.

Then:
 * $$\exists s \in S: \exists t_1, t_2 \in T: \left({s, t_1}\right) \in \mathcal R \and \left({s, t_2}\right) \in \mathcal R$$

By definition of inverse relation:
 * $$\exists s \in S: \exists t_1, t_2 \in T: \left({t_1, s}\right) \in \mathcal R^{-1} \and \left({t_2, s}\right) \in \mathcal R^{-1}$$

The composite of $\mathcal R^{-1}$ and $\mathcal R$ is defined as:


 * $$\mathcal R \circ \mathcal R^{-1} = \left\{{\left({x, z}\right) \in T \times T: \exists y \in S: \left({x, y}\right) \in \mathcal R^{-1} \and \left({y, z}\right) \in \mathcal R}\right\}$$

Thus:

$$ $$ $$

and so by definition of identity mapping $$R \circ \mathcal R^{-1} \ne I_T$$.

From this contradiction we deduce that $$\mathcal R$$ must be functional.

Hence the result.