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!

Please help :)

I've had an idea!

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. $\square$


enter image description here,

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.

  • $\begingroup$ Okay, yeah, I see the subcategory bit now; that's easy. Forget about the join-preservation. $\endgroup$ – Shaun Jul 21 '14 at 9:29
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    $\begingroup$ 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. $\endgroup$ – Martin Sleziak Jul 21 '14 at 9:59
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    $\begingroup$ 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. $\endgroup$ – Jakob Werner Jul 21 '14 at 11:20
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    $\begingroup$ Typo, of course I meant that Semigroups is not reflective in Monoids. $\endgroup$ – Jakob Werner Jul 21 '14 at 11:29
  • $\begingroup$ @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 :) $\endgroup$ – 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 $\iota$.

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  • $\begingroup$ (+1) That's been on the tip of my tongue for days now! Thank you! I upvoted this answer late last night but wanted to check it first before accepting. (Yes, it's correct! :D) $\endgroup$ – Shaun Jul 28 '14 at 7:54

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