An example where $R\circ R^{-1}=i_A$ but $R^{-1}\circ R\ne i_A$

Question

I'm looking for an example where R is a relation from set $A$ to $A$ and $R\circ R^{-1}=i_A$ (the composite of $R$ and inverse $R$ is equal to identity relation) but $R^{-1}\circ R\ne i_A$ (the composite of inverse $R$ and $R$ is not equal to identity relation).
For example A={1,2,3} and 1R2, 2R3 and 3R1 (but in this case both ways equals the identity relation).

Answer 1

xRy when xy = 0, a relation over the reals.

Answer 2

Let $A$ be the set of polynomials with real coefficients. Define $J:A\to A$ by $J(p)=\int_o^x{P(t)dt}$ and define $D:A\to A$ by $D(p)=p'(x).$ Then $D\circ J=id_A$ and $J\circ D \ne id_A$. You shouldn't have any trouble proving the former, or coming up with a counterexample to demonstrate the latter.

(Well, you shouldn't have any trouble if you've already had a calculus course. If you haven't, let me know, and I'll explain further.)