For some set called $A$, $\operatorname{id}A = \{(x,x): x ∈ A\}$. $R$ is a binary relation over $A$ $(R⊆A)$.
I need to prove that: $\operatorname{id}A ∘ R = R ∘ \operatorname{id}A = R$
I really do not understand what are the steps in order to prove that kind of a question. since it's such a trivial question I couldn't find a detailed proof anywhere.
Suppose $(x,y)\in R$. Now how does composition work? If $(a,b)\in R_1$ and $(b,c)\in R_2$, then $(a,c)\in R_2\circ R_1$. Thus, if you want to talk about $\operatorname{id}_A\circ R$, you need to take $(x,y)$, and look at pairs in $\operatorname{id}_A$ of the form $(y,\cdot)$. There is only one, namely $(y,y)$, therefore we have $(x,y)\in \operatorname{id}_A\circ R$.
That argument shows that $R\subseteq \operatorname{id}_A\circ R$. Show now that $\operatorname{id}_A\circ R\subseteq R\circ \operatorname{id}_A$, and also that $R\circ \operatorname{id}_A\subseteq R$, and you'll have equality for all three.
