# Does $R\subseteq A\times A$ being antisymmetric imply the same for $S$?

Given two sets $$A, B$$ let $$f:A\to B$$ be surjective and suppose $$R\subseteq A\times A$$ is an antisymmetric relation. Does it follow that $$S=\{(b,b')\in B\times B\ \vert \exists a,a'\in A: a R a', f(a) = b, f(a') = b' \}$$ is also antisymmetric?

Here's my work. If $$R=\emptyset$$ then $$S=\emptyset$$ and they're both antisymmetric.

So let $$R\neq\emptyset$$ be antisymmetric and suppose $$b,b'\in B$$ satisfy $$b S b'$$ and $$b' S b$$. Then there exist $$a_1,a_2,a_3,a_4\in A$$ such that $$a_1 R a_2, a_3R a_4$$ and $$f(a_1) = b = f(a_4), f(a_2)= b'=f(a_3).$$ Since in general we may assume that the $$a_i$$ are distinct and $$f$$ is not injective, $$b= b'$$ does not follow.

Is this correct, and enough? Or should I construct an explicit counterexample?

• You need an explicit counterexample. This amounts to saying "It might not hold if $f$ satisfies this condition," but as far as you know it holds nonetheless. You only showed that it could conceivably not hold. That's not really a proof of anything. – Matt Samuel Oct 26 '19 at 16:18
• My guess is that, when you wrote $S\times S$, you meant $B\times B$. – José Carlos Santos Oct 26 '19 at 16:20
• @MattSamuel: I see... well, thank you, I'll try to find one. Any hint is welcome. – Learner Oct 26 '19 at 16:21
• @JoséCarlosSantos Yes, thanks! – Learner Oct 26 '19 at 16:21

You've done an appropriate amount of work to be done, but you should express your work as a counterexample rather than as a condition that any counterexample would have. For instance, your $$A=\{a_1, a_2,a_3,a_4\}$$, $$B=\{b,b'\}$$, and so on. Then if you explicitly show that your order on $$A$$ is antisymmetric but your order on $$B$$ is not, then you're done.
Think about $$f=(x\mapsto|x|:\mathbb Z\to\mathbb N)$$.
• Thank you, I think I got it: with your $f$, $R\ =\ \le$ and $(a_1,a_2)=(-1,0), (a_3,a_4)=(0,1)$ one finds $f(a_1)=1=f(a_4), f(a_2)=0=f(a_3).$ Thus $1 S 0$ and $0 S 1$, but of course $0\ne1$. – Learner Oct 27 '19 at 7:52