3
$\begingroup$

In Semigroup Theory, Green's relations are everywhere. Their equivalence classes, for instance, on a given semigroup $S$ can tell one a lot about the structure of $S$.

There is some trivial sense in which a monoid can be viewed a category. Some people even argue that (at least small) categories are generalised monoids.

In light of the above, then, it seems natural to ask how to describe Green's relations using Category Theory, even if it is just for the category $\textbf{Monoids}$ and not $\textbf{Semigroups}$ or beyond.

If I had the time or the ability, I'd look into Lawvere theories a little deeper in my attempt to answer this question. I'm not entirely sure why. Perhaps it's because Universal Algebra is more visible in the structural approaches to Semigroup Theory I've seen.

I'm just curious :)

$\endgroup$
3
$\begingroup$

There is no need of category theory to define Green relations on categories. Let $C$ be a small category and let $0$ be a new element. Then $C \cup \{0\}$ is semigroup for the operation $*$ defined as follows: $$ x*y = \begin{cases} xy & \text{if $xy$ is defined in $C$}\\ 0 & \text{otherwise} \end{cases} $$ You can now just define the Green relations on this semigroup.

$\endgroup$
  • $\begingroup$ (+1) Thank you! I like this approach. However, I'm not sure yet whether to accept it as an answer. I'm interested in how category theory might be used to describe Green's relations, even if there's no need, not necessarily how Green's relations can be described on categories. I'm also wary of adjoining an element that acts like a zero, for as we know from semigroup theory, something like that can change things a little too much. I'm sorry. $\endgroup$ – Shaun Oct 5 '14 at 8:13
  • 1
    $\begingroup$ I introduced the zero to make the definition simpler, but it is not really used when it comes to Green relations. For instance, you can just say that $u\ \mathcal{R}\ v$ if there exist $x$ and $y$ such that $ux = v$ and $vy = u$. $\endgroup$ – J.-E. Pin Oct 5 '14 at 8:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.