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I'm doing preparaton problems for my exam and one of the first problems in the "composition of relations" section is this:

Prove: $$ (A \circ B)^{-1} = B^{-1} \circ A^{-1} $$

I know I need to prove 2 inclusions (L = Left side of the equation, R = right side of the equation):

$ L \subseteq R$ and $R \subseteq L$

After few first steps (in both cases) I'm stuck. My short "observations":

$ L \subseteq R$:

I know:

  1. $(x,y) \in (A \circ B) \Leftrightarrow \exists z: (x,z) \in A \land (z,y) \in B$, and
  2. $(x,y) \in A^{-1} \Leftrightarrow (y,x) \in A$.

So, let's take $(x,y) \in (A \circ B)^{-1}$. From (2): $(y,x)\in(A\circ B)$. From (1) and previous conclusion: $\exists z: (y,z) \in A \land (z,x) \in B$. But... what's now?

I tried to do something, but I don't know, how to solve this and similar problems. I showed some of my "work", so it's not just asking you to solve this problem for me. I'm also asking for explanation(s), how to think about this kind of problems.


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Your initial work is good. You seem to get lost, perhaps due to inexperience. Working step by step, and one at a time can be very helpful. Write different steps on different lines to organize your thoughts. E.g.:

  • Suppose that $(y,x)\in (A\circ B)^{-1}$, then $(x,y)\in A\circ B$.
  • Therefore there is $z$ such that $(x,z)\in A$ and $(z,y)\in B$.
  • Therefore $(z,x)\in A^{-1}$ and $(y,z)\in B^{-1}$.
  • Therefore $(y,x)\in B^{-1}\circ A^{-1}$.

The other direction is equally daunting if you work systematically.

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Thank you! I just needed to rearrange 3rd step to see the obvious: $(y,z)\in B^{-1}$ and $(z,x)\in A^{-1}$. It's the definition of composition of relations: $(y,x) \in B^{-1} \circ A^{-1}$, cause we know that required $z$ exists (2nd step in your solution). I write this all, because it can be useful for somebody who is as inexperienced as me and doesn't see simple connections at first. Anyway: thanks again, Asaf, your answer is very helpful! – golert Nov 11 '12 at 14:41
You're welcome. – Asaf Karagila Nov 11 '12 at 14:45
Solution for $R \subseteq L$: Suppose $(x,y) \in B^{-1} \circ A^{-1}$. That means: $\exists z: (x,z) \in B^{-1} \land (z,y) \in A^{-1}$. Therefore $(z,x) \in B \land (y,z) \in A$. We can rearrange this and write: $(y,z) \in A \land (z,x) \in B$, so it's easier to see that $(y,x) \in A \circ B$. So, $(x,y) \in (A\circ B)^{-1}$. I think it's ok (if not, please correct me), maybe it's useful for somebody who will have the same problem in the future. – golert Nov 11 '12 at 14:53

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