Find if the following equality is true for all relations.

Question

I need to prove or disprove that for all relations $A, B$ $\epsilon X^2$ that: $$(A\circ B)^{-1}=A^{-1}\circ B^{-1}.$$ I tried giving a counter example, but I'm not really sure if there even is one.

Answer 1

We have $(A\circ B)^{-1}=B^{-1}\circ A^{-1}$, and this would not usually equal $A^{-1}\circ B^{-1}$. So, just take a random example on a small set and hope for the best. Avoid special cases like $B=A$ or $B=A^{-1}$ or $B=\overline A$ or $B=\varnothing$. So say $$X=\{1,2\}\ ,\quad A=\{(1,1)\}\ ,\quad B=\{(1,2)\}\ .$$ Then $$A\circ B=\varnothing\quad\hbox{so}\quad (A\circ B)^{-1}=\varnothing$$ and $$A^{-1}\circ B^{-1}=\{(2,1)\}\ne (A\circ B)^{-1}\ .$$

Answer 2

It is not true if your relation $\circ $ is not commutative.

Take for instance $X=GL_2(\mathbb Z)$ and $\circ=\times$.

You will see that you have for $A,B\in GL_2(\mathbb Z)$:

$$(AB)^{—1}=B^{-1}A^{—1}$$

and not $(AB)^{—1}=A^{-1}B^{—1}$.