I am wondering how to view semigroups as categories. For example, we can easily view monoids as categories with a single object. Unfortunately, semigroups do not necessarily have identities, so the same view does not work with semigroups.

However, I did come across this idea where one adjoins a new element to a semigroup and treats it like its identity; effectively the semigroup gets promoted to a monoid. Of course, this new identity element must satisfy some additional properties to make sure that the original elements of the semigroup continue to behave as before. I am wondering what the additional properties might be.

  • 2
    $\begingroup$ If a semigroup $S$ has no identity, just add another element not in it, call it $E$ and assign the property $Ex=xE=x$, for all $x\in S$, plus $EE=E$. Nothing else is needed. There will be no invertible element other than $E$. The submonoids will be the subsemigroups of $S$, with the addition of $E$. $\endgroup$ – egreg Oct 6 '15 at 18:00
  • $\begingroup$ @egreg Hmm. I see. Thanks. A quick question: the notion of a "semigroup homomorphism" between two such "semigroups" must change accordingly to accommodate this right? Will it be a monoid homomorphism whose pre-image over the identity is just the identity (as in only the identity of one is mapped to the identity of the other)? $\endgroup$ – 0XLR Oct 6 '15 at 18:26
  • $\begingroup$ If you add an identity to all semigroups, including monoids, then there will be no difference at all. $\endgroup$ – egreg Oct 6 '15 at 19:04
  • $\begingroup$ @egreg "No difference at all" is a bit exaggerate. For instance, a semigroup $S$ is said to be locally trivial if, for all idempotents $e$, $eSe = e$. If $S$ is a monoid, this property implies that $S$ is trivial. $\endgroup$ – J.-E. Pin Oct 7 '15 at 7:29
  • $\begingroup$ @J.-E.Pin Thank you for your input. It seems that if I choose to view semigroups as monoids with an artificially added identity, I would need to change the conditions for local triviality to $e(S - \{1\})e = e$. Is there a cleaner way to view semigroups as categories then? Also, I have an unrelated question. Are you perhaps Jean-Éric Pin, author of the notes, "Mathematical Foundations of Automata Theory"? $\endgroup$ – 0XLR Oct 7 '15 at 20:45

The main reference on this question is [1]. You may look in particular to Section 1.2, p.22 (Foundations) and p. 95, where the definition of a semigroupoid is given.

The following is a quote from this book.

Given a semigroup S there are two popular ways to make S into a monoid: $S^I$ and $S^\bullet$ (this latter is often written $S^1$). The construction $S^I$ adjoins a new identity $I$ to $S$, even if $S$ already has an identity, whereas $S^\bullet$ adds an identity only when $S$ has no identity. The fi rst is the object part of a functor $\mathbf{Sgp} \to \mathbf{Mon}$, the second is not a functor; even better, $S \to S^I$ is the left adjoint of the forgetful functor from $\mathbf{Mon}$ to $\mathbf{Sgp}$. So (...) the construction $S \to S^I$ is the correct (or canonical) choice. This has the eff ect that for $G$ a group, $G^I$ is not a group. That's tough luck: from the categorical viewpoint, we cannot avoid this.

Now, another point of view is to consider semigroupoids instead of categories. Quoting [1] p. 95,

A semigroupoid is defi ned precisely like a category, but one relaxes the axiom demanding identities at each object. We can view a semigroup as a semigroupoid with a unique object.

[1] J. Rhodes, B. Steinberg, The $q$-theory of finite semigroups. Springer Monographs in Mathematics. Springer, New York, 2009. xxii+666 pp. ISBN: 978-0-387-09780-0

  • $\begingroup$ But if we view everything in the language of category theory (where a semigroup is then a semicategory with one element), then the forgetful functor from categories to semicategories has both a left and a right adjoint (the left being the one described here). The right is given by the Karoubi envelope. $\endgroup$ – Tobias Kildetoft Oct 8 '15 at 11:32
  • $\begingroup$ The karoubi envelope of a semigroup without idempotents is empty $\endgroup$ – Benjamin Steinberg Oct 8 '15 at 13:50
  • $\begingroup$ @BenjaminSteinberg As it should be if it is to give a right adjoint to the forgetful functor (as there are no morphisms from a monoid to a semigroup with no idempotents). $\endgroup$ – Tobias Kildetoft Oct 8 '15 at 17:34
  • $\begingroup$ But it means it won't serve the op's purposes. The karoubi envelope is only a good replacement for the semigroup when it has local units $\endgroup$ – Benjamin Steinberg Oct 8 '15 at 20:45
  • 1
    $\begingroup$ @ZeroXLR The "old" identity stops being an identity in the new semigroup, since it is not the identity for the "new" identity. $\endgroup$ – Tobias Kildetoft Oct 9 '15 at 6:47

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.