Is this another definition for inverse relation? We know that if $R\colon A\to B$, we can define the inverse relation as follows: $$R^{-1}=\{(y,x)\in B\times A\mid(x,y)\in R\}.$$ Now I want to know if this set, let's call it $R\:'^{-1}$, is the same as $R^{-1}$: $$R\:'^{-1}=\{(x,y)\in R\mid(y,x)\in B\times A\}.$$ I think $R^{-1}\neq R\:'^{-1}$ so I picked an example.
I tried with $A=\{1,2\}$, $B=\{3,4\}$, so $B\times A=\{(3,1),(3,2),(4,1),(4,2)\}$. Let's pick $R=\{(3,1),(4,1)\}$. Then, by definition, $$R^{-1}=\{(1,3),(1,4)\}.$$ But if we look at the definition of $R\:'^{-1}$, I think it says "All the elements of $(x,y)\in R$ such that the pair $(y,x)$ is in $B\times A$". Hence, $R\:'^{-1}=\emptyset$, because we know, for example, that $R$ has $x=3$ and $y=1$, so $(3,1)\in R$, but this proposition: $(y,x)=(1,3)\in B\times A$ is false. Thus $R^{-1}\neq R\:'^{-1}$.

Are my understanding of $R\:'^{-1}$ and counterexample correct?


We have $(y,x)\in B\times A$ whenever $(x,y)\in R\subseteq A\times B$. Hence your $R'^{-1}$ is just $R$ itself
