# A reflective subcategory of the category of inverse semigroups.

## The Question.

I'm reading Lawson's "Inverse Semigroups: The Theory of Partial Symmetries" and I've hit something I don't understand. It's claimed on page 34 of my copy that

The category of complete, infinitely distributive inverse semigroups together with join-preserving homomorphisms forms a subcategory of the category of inverse semigroups and homomorphisms; indeed, the former is a reflective subcategory of the latter.

Let's call them $$\mathbf{CompInfDist}_{\vee}$$ and $$\mathbf{InvSem}$$, respectively.

What exactly is the left adjoint of the inclusion functor that makes $$\mathbf{CompInfDist}_{\vee}$$ reflective in $$\mathbf{InvSem}$$?

## The Details.

This is technical stuff so let's have some definitions. Let $$S$$ be an inverse semigroup. Following Lawson . . .

Definition 1: The compatibility relation on $$S$$ is given by $$s\sim t\iff st^{-1}, s^{-1}t\in E(S),$$ where $$E(S)$$ is the set of idempotents of $$S$$.

Definition 2: A subset $$A$$ of $$S$$ is compatible if any pair of elements in $$A$$ are compatible.

The meet, $$a\wedge b$$, on $$S$$ for $$a, b\in S$$ is defined as the greatest lower bound of $$a$$ and $$b$$ with respect to the natural order on $$S$$; the join ($$\vee$$) is given dually. These extend to sets naturally.

Definition 3: We say $$S$$ is complete if every non-empty compatible subset of $$S$$ has a join.

Definition 4: We say $$S$$ is left infinitely distributive if, whenever $$A$$ is a non-empty subset of $$S$$ for which $$\bigvee A$$ exists, then $$\bigvee sA$$ exists for any element $$s\in S$$ and $$s\left(\bigvee A\right)=\bigvee sA$$. Then $$S$$ is infinitely distributive if it is both left and right infinitely distributive, where "right infinitely distributive" is defined analogously to left.

Now quoting MacLane,

Definition 5: A subcategory $$\mathcal{A}$$ of $$\mathcal B$$ is called reflective ($$*$$) in $$\mathcal B$$ when the inclusion functor $$K:\mathcal A\to\mathcal B$$ has a left adjoint $$F:\mathcal B\to\mathcal A$$.

## My Attempt.

I'm completely at a loss. I'm sorry. I've written out all the relevant definitions on my whiteboard, including peripheral, easy ones like "functor", "adjoint", "subcategory", etc., but I just don't see it.

Anyway, thank you for reading all of this!

Definition 6: A subset $$A$$ of $$S$$ is permissible if it is a compatible order ideal. The set of all permissible subsets of $$S$$ is denoted $$C(S)$$.

Lemma 1: $$C(S)$$ (under multiplication of subsets) is an object of $$\mathbf{CompInfDist}_{\vee}$$.

"Proof": This is Theorem 1.4.23 of Lawson's book. $$\square$$

$$\color{red}{\text{Perhaps}}$$ ,

where $$F=C'$$ is given by $$C'(S\stackrel{f}{\to}T)=C(S)\stackrel{Cf}{\longrightarrow}C(T)$$ for $$Cf: C(S)\to C(T)$$ given by $$A\mapsto f(A)$$. But this needs to satisfy $$\hom_{\mathbf{CompInfDist}_{\vee}}(C(S), Q)\cong_{\varphi_{(S, Q)}}\hom_{\mathbf{InvSem}}(S, K(Q)=Q)$$ for some natural bijection $$\varphi_{(S, Q)}$$.

Define $$\iota: S\to C(S)$$ by $$\iota(s)=[s]$$, where $$[s]$$ is the $$\sim$$-class of $$s$$.

Lemma 2: If $$\theta: S\to Q$$ is a homomorphism to an object $$Q$$ in $$\mathbf{CompInfDist}_{\vee}$$ then there exists a unique morphism $$\theta^*:C(S)\to Q$$ in $$\mathbf{CompInfDist}_{\vee}$$ given by $$\theta^*(A)=\bigvee\{\theta(a)\mid a\in A\}$$ such that $$\theta^*\iota=\theta$$.

"Proof": This is Theorem 1.4.24 of Lawson's book.$$\square$$

$$\color{red}{\text{Maybe}}\,\varphi^{-1}_{(S, Q)}(\theta)=\theta^*$$. But what's $$\varphi_{(S, Q)}$$ given by?

I'm not sure of the details :/

($$*$$) See the comments: I think this is the definition Lawson intended.

• Okay, yeah, I see the subcategory bit now; that's easy. Forget about the join-preservation. – Shaun Jul 21 '14 at 9:29
• According to Theorem 16.8 in Adamek-Herrlich-Strecker, if there is some factorization structure on a category (and fulfills some additional conditions), the epireflective subcategories are precisely the subcategories closed under (extremal) subobjects and products. I am not familiar with semigroups, but perhaps you are able to say whether your subcategory is closed under products, and extremal subobjects. – Martin Sleziak Jul 21 '14 at 9:59
• Mac Lane is not totally standard here. Usually reflective subcategories are assumed to be full. Otherwise, the concept is not very useful. For example, you wouldn't call $\mathbf {Semigroups}\subseteq\mathbf {Monoids}$ full, whereas $\mathbf {Monoids}\subseteq\mathbf {Groups}$ is reflective. – Jakob Werner Jul 21 '14 at 11:20
• Typo, of course I meant that Semigroups is not reflective in Monoids. – Jakob Werner Jul 21 '14 at 11:29
• @JakobWerner Thank you for the clarification. I suppose I should mention, then, that Lawson gave a reference to MacLane in the paragraph of the page-34 quote of mine above. Perhaps the non-standard definition was intended :) – Shaun Jul 21 '14 at 11:43

Your idea is absolutely right. The map $\varphi_{(S,Q)}$ maps a morphism $\psi \colon C(S) \rightarrow Q$ (in $\mathrm{CompInfDist}_\vee$) to the morphism $\psi \circ \iota_S$ (in $\mathrm{InvSem}$), where $\iota_S \colon S \hookrightarrow C(S)$ is the map you denoted as $\iota$.