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There are two operations $\otimes$ and $\odot$, which are anticommutative on a set X or Y. Additionally there is a function $f:X\to Y$ with the following property:

$f(r_X)=r_Y$ and $f(x\otimes y)=f(x)\odot f(y)$, for all $x,y\in X$. (By the way, has that property a proper name?)

Then, why is it true that $xRy:\Leftrightarrow f(x\otimes y)=r_Y$ is a equivalence relation on X? In detail my question relates to the part of the proof, where the symmetric property is verified.

Assume if $x,y\in X$ then one has $f(x\otimes y)=r_Y$. Hence by the anticommutative property of $\otimes$ and $f(r_X)=r_Y$ we can conclude $x\otimes y=r_X \Leftrightarrow x=y$. Set $a:=y$ and $b:=x$ then we can conclude $a\otimes b=r_X$, further we get $f(a\otimes b)=r_Y=f(y\otimes x)$. This is equivalent to $(y,x)\in R_\Box$

An other way could be:

Assume if $x,y\in X$ then one has $f(x\otimes y)=r_Y$. Hence by the anticommutative property of $\otimes$ and $f(r_X)=r_Y$ we can conclude $\underbrace{(x\otimes y)}_{r_X}\otimes (y\otimes x)=r_X$. Again with the anticommutative property of $\otimes$ then one has $r_X\otimes\underbrace{(y\otimes x)\otimes r_X}_{(y\otimes x)}=r_X\otimes r_X$, which is equivalent to $(y\otimes x)=r_X$. This implies $r_Y=f(y\otimes x)$, which is equivalent to $(y,x)\in R_\Box$

If I'm not wrong is there a smarter way to proof the symmetric property?

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StellaMaris
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1 Answers1

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Your question looks similar to exercise 5 in the linked chapter of this book. I add the definition of an anti- or not commutative opertaion, because we need it for the argumentation.

An anticommutative operation on a set X to be a function $\cdot:X\times X\rightarrow X$ satisfying two properties:
(i) Existence of right identity: $\exists r\in X:x\cdot r=x$ for all $x\in X$
(ii) $x\cdot y=r\iff(x\cdot y)\cdot(y\cdot x)=r\iff x=y$ for all $x,y\in X$

If you want to know if your relation $R\subset X\times X$ has symmetric property, where $R$ is defined as $R=\{(x,y)\in X\times X | f(x\otimes y)=r_Y\}$, you need to check if $(x,y)\in R$, then $(y,x)\in R$ as well.

For any $(x,y)\in R$ one may infer: $$f(x\otimes y)=f(x)\odot f(y)=r_Y.\hspace{3cm} (1)$$

For $\odot$-operation we know from (ii) that: $$ f(x)\odot f(y)=r_Y \implies f(x)=f(y) \hspace{3cm} \forall f(x),f(y)\ \ (2)$$

With (1) we can conclude from (2) that $f(x)=f(y)$. If this is true, $r_y=f(x)\odot f(y)=f(y)\odot f(x)=f(y\otimes x)$ has to be true. That means $(y,x)\in R$.

...(By the way, has that property a proper name?)...

Maybe faithful or action faithful

Hölderlin
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