Let $A\neq\emptyset$ and $\mathcal{R}\subset A\times A$ a binary relation on $A$ such that the domain $D(A):=\{x\in A\mid \exists y\in A \text{ such that }(x,y)\in\mathcal{R}\}=A$ and $\mathcal{R}\circ\mathcal{R}^{-1}\circ \mathcal {R}=\mathcal{R}$. The exercise requires to prove that $\mathcal{R}\circ\mathcal{R}^{-1}$ and $\mathcal{R}^{-1}\circ \mathcal {R}$ are equivalence relations.
I do not manage to prove that $\mathcal{R}\circ \mathcal {R}^{-1}$ is symmetric. For $\mathcal{R}^{-1}\circ \mathcal {R}$ it seemed quite simple: Let $x\in A$; since the domain of the relation is $A$, it follows that $\exists y\in A$ such that $(x,y)\in \mathcal{R}$. It follows that $(y,x)\in\mathcal{R}$ and therefore $(x,x)\in \mathcal{R}^{-1}\circ \mathcal {R}$. For the other relation I cannot seem to prove it.