# Viewing Semigroups as Categories?

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.

• 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 $E$. – egreg Oct 6 '15 at 18:00
• @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)? – 0XLR Oct 6 '15 at 18:26
• If you add an identity to all semigroups, including monoids, then there will be no difference at all. – egreg Oct 6 '15 at 19:04
• @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. – J.-E. Pin Oct 7 '15 at 7:29
• @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"? – 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

• 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. – Tobias Kildetoft Oct 8 '15 at 11:32
• The karoubi envelope of a semigroup without idempotents is empty – Benjamin Steinberg Oct 8 '15 at 13:50
• @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). – Tobias Kildetoft Oct 8 '15 at 17:34
• 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 – Benjamin Steinberg Oct 8 '15 at 20:45
• @ZeroXLR The "old" identity stops being an identity in the new semigroup, since it is not the identity for the "new" identity. – Tobias Kildetoft Oct 9 '15 at 6:47