An equivalence functor between local vector lattices and vector lattices

We call a local vector lattice any vector lattice with a distinguished positive strong unit and having exactly one maximal ideal (its radical). We provide a short study of local vector lattices. In this regards, some characterizations of local vector lattices are given. For instance, we prove that a vector lattice with a distinguished strong unit is local if and only if it is clean with non no-trivial components. Nevertheless, our main purpose is to prove, via what we call the radical functor, that the category of all vector lattices and lattice homomorphisms is equivalent to the category of local vectors lattices and unital (i.e., unit preserving) lattice homomorphisms.


Introduction
All vector lattices we consider in this paper are assumed to be real.Let L and M be vector lattices.Recall that a linear map L The category whose objects are all vector lattices and morphisms are lattice homomorphisms is denoted by VL.Thus, by a VL-object we mean a vector lattice and any lattice homomorphism between two VL-objects is referred to as a VL-morphism.Now, a vector subspace I of a VL-object E is called an (order ) ideal in E if I contains with any x all y such that |y| ≤ |x|.Clearly, if x ∈ E then the set I x = {y ∈ E : |y| ≤ n |x| , for some n = 1, 2, ...} is an ideal in E. Actually, I x is the smallest ideal in E containing x.An element x ∈ E for which I x = E is called a strong (order ) unit in E. As usual, an ideal I in E is said to be proper if I = E.By the way, a proper ideal in E contains no strong units.A proper ideal M in E is said to be maximal if M is not contained in any other proper ideal in E. We call a unital vector lattice any VL-object along with a distinguished positive strong unit.Moreover, if E and F are unital vector lattices with distinguished strong units e and f , respectively, then any VL-morphism E ω → F for which the equality ω (e) = f holds is said to be unital.It is well-known that any unital vector lattice E has maximal ideals (see [7,Theorem 27.4]).It could happen that E has exactly one maximal ideal.In this situation, we call E a local vector lattice (see [8] for the corresponding concept in Ring Theory).The subcategory of VL with objects all local vector lattices and morphisms unital lattice homomorphisms is denoted by LVL.In other words, an LVLobject is a local vector lattice and an LVL-morphism is a unital lattice homomorphism between two local vector lattices.The main purpose of this note is to prove that the categories VL and LVL turn out to be equivalent.The equivalence functor R : VL → LVL in question is defined as follows.
As proved in [7,Theorem 27.4], any LVL-object E has maximal ideals.Then, the intersection of all maximal ideals in E is the radical of E, which we denote by Rad (E).For every LVL-morphism Actually, ω 0 is well-defined because, as we shall see later, ω sends Rad (E) to Rad (F ).Roughly speaking, ω 0 is the restriction of ω to Rad (E).Then, R is the functor that takes each LVL-object E to the VL-object Rad (E), It will turn out that R is the functor we are looking for.
Finally, we point out that the standard monograph [7] is used as a unique source of unexplained terminology and notations on Vector Lattices.But for Category Theory, the reader can consult the text [6].

Preliminaries on local vector lattices
As usual, the symbol R is used to denote the totally-ordered field of all real numbers.As mentioned before in the case of local vector lattices, any unital vector lattice E has maximal ideals (see Theorem 27.4 in [7]) and the radical of E (i.e., the intersection of all maximal ideals in E) is denoted by Rad (E).It turns out that Rad (E) coincides with the ideal E 0 of all infinitely small elements of E (see [7,Theorem 27.5 ]).Recall here that an element x ∈ E is said to be infinitely small if there exists y ∈ E such that holds, where Re is the line generated by the vector e.Consequently, if E has a unique maximal ideal, then E = Re ⊕ Rad (E).Conversely, if E = Re ⊕ Rad (E) then Rad (E) is the only maximal ideal in E. Indeed, let M be a maximal ideal in E and pick x ∈ M .Then there exists r ∈ R and It follows that r = 0 (otherwise, M would contains a strong unit which is impossible).Therefore, and so We derive directly that Rad (E) is the unique maximal ideal in E. We call E a local vector lattice provided one the following equivalent conditions hold.
(1) E has a unique maximal ideal.
Thus, for any element x in the local vector lattice E there exist a unique real number r (x) and a unique element ρ (x) ∈ Rad (E) such that We call r (x) e the real component of x and ρ (x) its infinitely small component.By the way, the maps r and ρ that take each x ∈ E to r (x) and ρ (x), respectively, are linear.But one might still wonder what a local vector lattice looks like.Some elements are clarified next.
Lemma 2.1.Let E be a local vector lattice and x ∈ E. Then the following hold.
(ii) Suppose that x / ∈ Rad (E) or, equivalently, r (x) = 0. Using (i) and the equality r (−x) = −r (x), we infer that Summarizing, we get the equality |x| = sign (r (x)) x and (ii) follows. (iii This completes the proof of the lemma. The following characterization of local vector lattices may well not have been quite on the agenda, but we think that it could be of interest to some.Let E be vector lattice with a distinguished strong unit e > 0. We call a component of E any element p ∈ E such that p ∧ (e − p) = 0.
Notice that 0 and e are components in E. The set of all components in E is denoted by C (E).Clearly, 0 ≤ p ≤ e and e − p ∈ C (E) , for all p ∈ C (E) .
More about components can be found in [1,Section 1.4].Following the recent work [5] (see also [4]), we say that E is clean if each element of E can be written as a sum of a strong unit and a component of E. In other words, E is clean if and only if, for every x ∈ E there exist a strong unit u of E and p ∈ C (E) such that x = u + p.We are in position now to give the characterization we were talking about.
Theorem 2.2.A vector lattice E with with a strong unit e > 0 is local if and only if E is clean and C (E) = {0, e}.
Proof.Necessity.Choose p ∈ C (E) and assume that 0 < p < e.This assumption together with the equality p ∧ (1 − p) = 0 yields that p is not a strong unit in E (otherwise, we would get e − p = 0 and so p = e).Since Rad (E) is the unique maximal ideal in E, Theorem 27.4 in [7] shows that p ∈ Rad (E).Analogously, e − p ∈ Rad (E) and so But Rad (E) is a maximal ideal in E and thus cannot contain strong units.This contradiction allows us to conclude that p ∈ {0, e} and so C (E) = {0, e}.Now, we claim that E is clean.To this end, we prove that if x ∈ E then x = u + p for some strong unit u in E and p ∈ {0, e}.If x itself is a strong unit in E, then take u = x and p = 0. So, suppose that x is not a strong unit in E. Again by Theorem 27.4 in [7], we infer that x ∈ Rad (E).Assume that e − x is not a strong unit in E. Similarly, we would obtain that e − x ∈ Rad (E) which leads to the contradiction Therefore, e − x is a strong unit in E. It suffices thus to put u = e − x and p = e.This means that E is clean and proves Necessity.
Sufficiency.Assume that E is clean and C (E) = {0, e}.Let M and N be two maximal ideals in E. We claim that M = N .To this end, we shall argue by contradiction.Without loss of generality, suppose that there exists x ∈ M such that x / ∈ N .So, there is r ∈ R for which r = 0 and x − re ∈ N (see [7,Theorem 27.3]).By (iii), we can find a strong unit u in E and p ∈ C (E) = {0, e} such that If p = 0 then ru = x ∈ M , which contradicts the maximality of M .Analogously, if p = e then ru = x − re ∈ N , contradicting here the maximality of N .We deduce that M = N and then E has only one maximal ideal.We can conclude that E is local, as desired.
The following (and last) result in this section will be of much greater use in the next section, in which categories are involved.For this reason, we are once again using the Category language.Recall from the introduction that VL denotes the category of all vector lattices and lattice homomorphisms, while LVL is the used to denote the subcategory of VL whose objects are local vector lattices and morphisms are unital lattice homomorphisms.For convenience, recall that if E and F are LVL-objects with distinguished positive strong units e and f , respectively, then a LVL-morphism E ω → F is a VL-morphism for which ω (e) = f .For instance, it follows from the third assertion in Lemma 2.1 that the map r : E → R that takes each element x ∈ E to the real number r (x) is an LVL-morphism.Notice by the way that R is an LVL-object with Rad (R) = {0}.As previously pointed out, we end this section with a lemma that will often come in handy in the next section.
Proof.Since the radical of a unital vector lattice coincides with the ideal of all infinitely small elements, it suffices to prove that if an element x ∈ E is infinitely small, then so is ω (x) ∈ F .So, let x be an infinitely small element of E. Therefore, there exists y ∈ E such that It follows that ω (x) is an infinitely small element of F , leading to the desired result.

The radical functor
Notwithstanding its simpleness, Lemma 2.3 allow us to introduce a functor from LVL to VL.Indeed, for every LVL-morphism E ω → F , we define a VL-morphism Rad (E) Roughly speaking, ω 0 is the restriction of ω to Rad (E).Then, let R be the function that takes each LVL-object E to the VL-object Rad (E), and each We can readily check that R is a functor from LVL to VL.We call R : LVL → VL the radical functor.We begin our investigation with the following first property of the radical functor.We emphasize that we shall keep the same notations as previously used in the preceding sections.
We claim that ω = ψ.To this end, pick x ∈ E and observe that This completes the proof of the lemma.
Next, we give the second fundamental property of the radical functor.
Proof.Let E, F be two LVL-objects, and choose a VL-morphism Rad (E) ϕ → Rad (F ).We claim that there exists an LVL-morphism E ω → F such that To this end, let e and f denote the distinguished positive strong units of E and F , respectively.Define ω : E → F by putting Observe that ϕ (ρ (x)) ∈ Rad (F ) (where we use Lemma 2.3).Furthermore, it follows readily from the direct sum that ω is linear.Also, ω (e) = f because r (e) = 1 and ρ (e) = 0.Moreover, if x ∈ Rad (E) then r (x) = 0 and thus It remains to show that ω is a lattice homomorphism.Let x ∈ E and observe that It follows quickly, again from Lemma 2.3, that Hence, by linearity and Lemma 2.1 (iii), we derive that Thus, ω is a lattice homomorphism the proof is complete.
To conclude that the radical functor is an equivalence, it remains to show that it is isomorphism-dense.More precisely, we have to show that for every VL-morphism L ϕ → M , there exists an LVL-morphism It turns out that this property is quite involved and requires much more work than the properties of being faithful and full.In this prospect, further backgrounds are needed.Let L be a VL-object.The Cartesian product R × L is a real vector space with respect to the coordinatewise addition and scalar multiplication.Moreover, it is clear that L can be identified with the vector subspace {0} × L of R × L. Henceforth, we shall consider L as a vector subspace of R × L. Analogously, identifying R with R × {0}, we can assume that R is a vector subspace of R × L. Thus, we have the direct sum Consequently, each vector x ∈ R ⊕ L can be written uniquely as a sum of a scalar Re (x) ∈ R and a vector L (x) ∈ L.Then, we have Also, we can perform a simple calculation to see that Re (rx + y) = r Re (x) + Re (y) , for all r ∈ R and x, y ∈ R ⊕ L (3.1) and We have gathered now all the ingredient we need to prove the following theorem.
Theorem 3.3.Let L be a VL-object.Then R ⊕ L is an LVL-object with radical L and positive strong unit 1.
Proof.We shall keep the above notations.Let K denote the set of all x ∈ R ⊕ L such that, either Re (x) > 0, or Re (x) = 0 and x ∈ L + .First, we claim that K is a cone in R ⊕ L (see Definition 1.2 in [3]).To this end, notice that 0 ≤ Re (x) , for all x ∈ K.
Accordingly, Re (x) = 0 and so ±x ∈ L + .This yields that x = 0 and finally that K is a cone in R ⊕ L, as required.In other words, R ⊕ L is an ordered vector space with K as positive cone (see Theorem 11.4 in [7]).Henceforth, we give up the symbol K and we use instead the classical notation (R ⊕ L) + .In the next step, we prove that the order vector space R⊕L is a vector lattice.Actually, we shall see that any element in R ⊕ L has an absolute value in R ⊕ L. Hence, let x ∈ R ⊕ L and define y ∈ R ⊕ L by Observe that if Re (x) = 0 then y = |Re (x)| + sign (Re (x)) L (x) .
We infer that y ∈ (R ⊕ L) + .Moreover, and if Re (x) < 0 then Analogously, we may show that y We derive that This means that the inequality |x| ≤ n holds in R ⊕ L. It follows that 1 is a strong unit in R ⊕ L, as required.The last step in the proof is to show that L is the unique maximal ideal (where we use (3.1) and (3.3)).This yields that Re (y) = 0 and so y ∈ L. We infer that L is an ideal in R ⊕ L, as desired.On the other hand, choose a maximal ideal M and pick x ∈ M .We claim that Re (x) = 0. To this end, we shall argue by contradiction assuming that Re (x) > 0 (if Re (x) < 0 then we can work with −x).Observe that 2 Re (x) But then x would be a strong unit in R ⊕ L, which contradicts the fact that M is a maximal ideal.Therefore, Re (x) = 0 and thus M is contained in L. By maximality, we obtain that L = M .This yields that L is the unique maximal ideal in R ⊕ L and completes the proof of the theorem.
As an application of Theorem 3.3, we get the following property of the radical functor R : LVL → VL.It turns out that ω is an LVL-morphism.Indeed, it is clear that ω (1) = 1.Moreover, the linearity of ω follows straightforwardly from (3.1) and (3.2).Also, if x ∈ L then Re (x) = 0 and so ω (x) = ϕ (L (x)) = ϕ (x) .We conclude that the radical functor R : LVL → VL is full.
Combining Lemma 3.1, Lemma 3.2, and Corollary 3.4, we get straightforwardly the following theorem, which is the main (and the last) result of the paper.

Lemma 3 . 1 .
The radical functor R : LVL → VL is faithful.Proof.Let E, F be two LVL-objects and let e, f denote the distinguished positive strong units of E and F , respectively.Pick two LVL-morphisms E ω → F and E ψ → F such that Consequently, y is an upper bounded in R ⊕ L of the pair {−x, x}.Choose another upper bound z of the pair {−x, x} in R ⊕ L. We derive that Re (z + x) ≥ 0 and Re (x − z) ≥ 0. It follows from (3.1) that Re (z) ≥ |Re (x)|.If Re (x) = 0 then y = sign (Re (x)) x = ±x ≤ z.Also, if Re (z) = 0 then Re (x) = 0 and so x, z ∈ L. Accordingly, the inequalities z ≥ ±x holds in L which gives z ≥ |x| = y.Finally, if Re (x) = 0 and Re (z) > 0 then, by (3.1), Re (z − y) = Re (z − |x|) = Re (z) > 0Accordingly, z − y ∈ (R ⊕ L) + and finally y is precisely the absolute value of x in R ⊕ L. Summarizing, R ⊕ L is a vector lattice and the absolute value in R ⊕ L is given by|x| =    |L (x)| if Re (x) = 0 sign (Re (x)) x = |Re (x)| + sign (Re (x)) L (x) if Re (x) = 0.In particular,Re (|x|) = |Re (x)| , for all x ∈ R ⊕ L (3.3) and L (|x|) = sign (Re (x)) L (x) , for all x ∈ R ⊕ L with Re (x) = 0. (3.4)At this point, observe that 1 ∈ (R ⊕ L) + .We claim that 1 is a strong unit in R ⊕ L. To do this, pick x ∈ R ⊕ L and choose n ∈ {1, 2, ...} such that n > |Re (x)|.Hence, by (3.1), Re (n − x) = n − Re (x) > 0 and Re (n + x) = n + Re (x) > 0.

Corollary 3 . 4 .
The radical functor R :LVL → VL is isomorphic-dense.Proof.Let L ϕ → M be a VL-morphism.Put E = R ⊕ L and F = R ⊕ L.From Theorem 3.3 it follows that E is an LVL-object with 1 as a positive strong unit and Rad (E) = L. Similarly, F is an LVL-object with 1 as a positive strong unit and Rad (F ) = M .Now, since ϕ (L (x)) ∈ M, for all x ∈ L, we can define a map E ω → F by putting ω (x) = Re (x) + ϕ (L (x)) , for all x ∈ E.