-1

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).

Avner Levy
  • 99
  • $R^{-1}$ is usually a two-sided inverse, I think. You should probably say you want an example where $R\circ S= id$ but $S\circ R\ne id$ – saulspatz Apr 10 '18 at 03:09
  • Hint: Consider $R$ to be a surjective (but not injective) map on $A$. Then, it has a right inverse but no left inverse. – Prasun Biswas Apr 10 '18 at 03:15
  • You should define $R^{-1}$. If $R \subset A \times A$ is a set of ordered pairs is $R^{-1}$ the set of reversed ordered pairs? If not, what is it? – Ross Millikan Apr 10 '18 at 03:17

2 Answers2

0

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

William Elliot
  • 17,598
0

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.)

saulspatz
  • 53,131