invertible relation

**xixihaha** · March 14th 2010, 03:44 PM

hi
my teacher wants us to think a conjecture for the invertible relation and then prove it, but I have absolutely no idea how to find that conjecture. Any help is appreciated.

$R,S \subset X \times X$ , we define $R \bullet S \text{ to be the set:} (x,y) \in X \times X , \text{s.t there exsits one and only one}$ $z \in X : (x,z) \in S,(z,y) \in R$

like a relation $\mathcal{R} \subset X \times X$ is invertible with respect to $\bullet$ if and only if (what condition does R has to satisfy?)
any idea?

cheers.

**xixihaha** · March 15th 2010, 03:51 AM

any idea?

**emakarov** · March 15th 2010, 07:21 AM

, we define

This is not a regular composition of relations. In the standard definition of composition there is no restriction that there is at most one $z$ . Is this a point of the exercise or a typo?

**emakarov** · March 15th 2010, 07:36 AM

Also, by $R$ being invertible you mean that there exists an $S$ such that $S\bullet R=Id$ or ( $R\bullet S=Id$ ) where $Id$ is the diagonal relation?

Probably this has something to do with surjection and injection. It is well-known that a function is an injection iff it has a left inverse, and is a surjection iff it has a right inverse (I hope I did not mix it up). Maybe something similar is expected here.

**xixihaha** · March 15th 2010, 10:42 AM

Originally Posted by emakarov

This is not a regular composition of relations. In the standard definition of composition there is no restriction that there is at most one $z$ . Is this a point of the exercise or a typo?

yea this is the exericese that our lecturer gives us.

**xixihaha** · March 15th 2010, 10:46 AM

Originally Posted by emakarov

Also, by $R$ being invertible you mean that there exists an $S$ such that $S\bullet R=Id$ or ( $R\bullet S=Id$ ) where $Id$ is the diagonal relation?

Probably this has something to do with surjection and injection. It is well-known that a function is an injection iff it has a left inverse, and is a surjection iff it has a right inverse (I hope I did not mix it up). Maybe something similar is expected here.

$S\bullet R= R\bullet S=Id$ ) where $Id$ is the diagonal relation. yes.
any idea what the conjecture is??

my thought is $\forall y \in x$ there exists one and only one $x \in X, (x,y) \in R$
is this correct?

**emakarov** · March 16th 2010, 06:09 AM

my thought is

there exists one and only one

I think this not correct. Let X = {1, 2} and let R = {(1, 1), (1, 2), (2, 2)}. Then R^2 = Id.

I don't know what the answer is, it requires some thinking and considering examples.

**xixihaha** · March 16th 2010, 02:27 PM

Originally Posted by emakarov

I think this not correct. Let X = {1, 2} and let R = {(1, 1), (1, 2), (2, 2)}. Then R^2 = Id.

I don't know what the answer is, it requires some thinking and considering examples.

i dont know either, but anyway thanks for your attention.

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