An extra condition for composition of relations which are not necessarily functions Given two binary relations $\mathcal R$ and $\mathcal S$ over $A \times B$ and $B \times C$,then their composition relation is denoted by $\mathcal S \circ \mathcal R$ and is defined as:
$$\mathcal S \circ \mathcal R:=\left\{(a,c) \mid \exists \;b \in B: a\mathcal R b \wedge b\mathcal S c\right\}$$
The converse (inverse) of composition of relations is given by:
$$(\mathcal S \circ \mathcal R)^T=\mathcal  R^T \circ \mathcal S^T$$
The proof is as follows:
$$
\begin{align}
\\
&amp;(c,a) \in (\mathcal S \circ \mathcal R)^T\\
&amp;\iff (a,c) \in (\mathcal S \circ \mathcal R)\tag{1}\\
&amp;\iff\exists \;b \in B: a\mathcal R b \wedge b\mathcal S c \tag{2}\\
&amp;\iff\exists \;b \in B:   c\mathcal S^T b \wedge b\mathcal R^T a\tag{3}\\
&amp;\iff(c,a) \in   \mathcal R^T \circ  \mathcal S^T \tag{4}\\
&amp;\iff (\mathcal S \circ \mathcal R)^T=\mathcal R^T \circ  \mathcal S^T\tag{5}
\end{align}
$$
Explanation:
$(1)\;,(3)$: Definition of converse relations
$(2)\;,(4)$: Definition of composition of relations
$(5)$: Definition of equality of sets

Proofwiki gives another condition for $(\mathcal S \circ \mathcal R)^T$ to exist:

From the definition of an inverse relation, we have:


$\mathsf{Domain}(\mathcal R_2)=\mathsf{Codomain}(\mathcal R^T_2)$


$\mathsf{Codomain}(\mathcal R_1)=\mathsf{Domain}(\mathcal R^T_1)$


So we confirm that $\mathcal R^T_1∘\mathcal R^T_2$ is defined.

I don't understand why do we need to check the existence of $(\mathcal S \circ \mathcal R)^T$,since we are not talking about functions and their left-totality property,so it's not the case and we don't need such a restriction,unless the relations are functions.
Am I right?If my argument is not true then please explain.
I don't think this is an extra condition. We known that a necessary condition for $\mathcal S \circ \mathcal R$ to be well-defined is that $\text{Codomain } \mathcal R = \text{Domain } \mathcal S$ - this is the set that you refer to as $B$ in you definition. The author is simply pointing out that this means that $\text{Codomain } \mathcal S^T = \text{Domain } \mathcal R^T$ and so $\mathcal R^T \circ \mathcal S^T$ is also well-defined.
&quot;We confirm&quot; is short for &quot;we can confirm&quot; or &quot;it is clear that&quot;, not an instruction that &quot;we must confirm&quot;.
I don't think this is an extra condition. We known that a necessary condition for $\mathcal S \circ \mathcal R$ to be well-defined is that $\text{Codomain } \mathcal R = \text{Domain } \mathcal S$ - this is the set that you refer to as $B$ in you definition. The author is simply pointing out that this means that $\text{Codomain } \mathcal S^T = \text{Domain } \mathcal R^T$ and so $\mathcal R^T \circ \mathcal S^T$ is also well-defined.
&quot;We confirm&quot; is short for &quot;we can confirm&quot; or &quot;it is clear that&quot;, not an instruction that &quot;we must confirm&quot;.