# a category associated with an inverse semigroup

Let $S$ be an inverse semigroup. Define a category $C(S)$ as follows:

• the objects of $C(S)$ are the elements of $S$
• for any $a,b,e\in S$ let $e\colon a\to b$ be a morphism of $C(S)$ iff $aa^{-1}eb^{-1}b=e$
• if $e\colon a\to b$ and $f\colon b\to c$ are morphisms, let $f\circ e\equiv eb^{-1}f\colon a\to c$ (this is a valid morphism because $eb^{-1}f=(aa^{-1}eb^{-1}b)b^{-1}(bb^{-1}fc^{-1}c)=aa^{-1}(eb^{-1}f)c^{-1}c$)

(I would much prefer to write $e\circ f$ for the composition but am bowing to convention.) Easy to see that composition is associative. Identities are $a\colon a\to a$. So it's a category. (I'm tempted to call it the overlap category.) This construction occurred to me a few days ago and I'm puzzled because I don't remember coming across it in the literature. But I'm not well versed with inverse semigroups. Anyone seen this before?

• It would be more natural if you called your $e$ as $e:a \rightarrow b^{-1}$. But then you have a problem with composition with $e:a \rightarrow b$ and $f:b \rightarrow c$, because in the ranges you flawly" forgot $^{-1}$. So you put a $b^{-1}$ between the composition of $e$ and $f$ to repair this. Maybe this works also more generally for other situations, where the objects are functions. – hänsel Mar 15 '18 at 1:21

I think this is isomorphic to what Alfredo and I call the Schutzenberger category of a semigroup in this paper (arXiv link) .

We define it for semigroups in general but it should boil down to what you wrote for inverse semigroups. We however would say your arrow e goes from b to a. So I guess maybe it is the opposite category but for an inverse semigroup it doesn't matter.

For an inverse semigroup, or more generally a von Neumann regular semigroup, this category is equivalent to the Karoubi envelope (aka idempotent splitting or Cauchy completion). For non-regular semigroups it is more interesting.

The journal version is here (Springerlink). Reference: (A. Costa and B.Steinberg, The Schützenberger category of a semigroup, Semigroup Forum (2015) 91(3) 543–559)

• That's fantastic, thanks! I'll have to dig in but am already pretty sure this is the same thing because I did the other day find that equivalent objects were given by the D relation. Also was surprised to see that this work was related to study of flow equivalence. The project that I'm working on is also related to flow equivalence! Will report back after I read your papers (which may take a while). – David Hillman Mar 15 '18 at 23:16
• The translation is $aa^{-1}eb^{-1}b=e$ iff e is in $aS\cap Sb$ and if $e,f$ are as in the question then $e=eb^{-1}b$ and $f=bb^{-1}f$ and our multiplication rule is to remove one of the two $b$'s and multiply which gives $eb^{-1}bb^{-1}f=eb^{-1}f$. So we have exactly the same category except we oriented our arrows in opposite directions. – Benjamin Steinberg Mar 16 '18 at 11:24
• BTW I'd be curious if you can find applications to flow equivalence beyond what we did. – Benjamin Steinberg Mar 20 '18 at 18:33

I have not a name for it, but as already indicated in my comment, your categroy $C$ is isomorphic to the following more clear category $D$:

Objects of $D$ are $S$. Morphisms from $a \in S$ to $b \in S$ are $e \in S$ with $a a^{-1} e b b^{-1} = e$. Composition of morphisms is just ordinary product in $S$.

Then $F:C \rightarrow D: F(e:a \rightarrow b ) := (e b^{-1} :a \rightarrow b)$ is the isomorphism between $C$ and $D$.

• This is connected to why the category is equivalent to the Karoubi envelope in the regular case but you can't get to ordinary composition in the non-regular case where the automorphism groups are the classical Schutzenberger groups, whence the name, which don't necessarily embed in the semigroup. – Benjamin Steinberg Mar 15 '18 at 12:55
• Thanks for that. I would add though that beauty is in the eye of the beholder! For instance, I like $C$ because we have inverse acting as a period-2 contravariant functor from $C$ to itself. Namely: if $e\colon a\to b$ (because $e=aa^{-1}eb^{-1}b$) then also $e^{-1}\colon b^{-1}\to a^{-1}$ (because $e^{-1}=b^{-1}be^{-1}aa^{-1}$). – David Hillman Mar 15 '18 at 23:10