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B. de Pagter and A.W. Wickstead in [7]. The free Banach lattice generated by a set A, denoted by FBL(A), was further extended in [5] to define the concept of free Banach lattice generated by a Banach space E, denoted by FBL[E]. This is a Banach lattice together with a linear isometry δE : E → FBL[E] such that for every Banach lattice X and every bounded linear operator T : E → X there is a unique lattice homomorphism Tˆ : FBL[E] → X such that Tˆ ◦ δE = T and moreover kTˆk = kT k. Existence and uniqueness up to Banach lattice isometries of free Banach lattices were proved in [5] (also in [7] for the case of FBL(A)), and an explicit description was provided in [5, Theorem 2.5], as follows. Let E be a Banach space. Given a positively homogeneous function f : E∗ → R (i.e., f(λx∗) = λf(x∗) for every λ > 0 and every x∗ ∈ E∗), we consider

n n ∗ N ∗ ∗ ∗ ∗ kfkFBL[E] := sup |f(xi )| : n ∈ , x1,...,xn ∈ E , sup |xi (x)|≤ 1 ∈ [0, ∞]. x B (i=1 ∈ E i=1 ) X X ∗ We denote by H0[E] the linear space of all positively homogeneous functions f : E → R such that kfkFBL[E] < ∞, which becomes a Banach lattice when equipped with the norm k·kFBL[E] and the pointwise order. Then the Banach lattice FBL[E] is the sublattice of H0[E] generated by the set {δE(x): x ∈ E}, where δE(x) denotes the evaluation map defined by

∗ ∗ ∗ ∗ ∗ δE(x): E → R, δE(x)(x ) := x (x) for all x ∈ E .

These evaluations form the natural copy of E inside FBL[E], via the linear isometry δE : E → FBL[E]. It is worth noticing that in the case of the underlying Banach space E being a Banach lattice, its lattice structure seems to be “forgotten” when it embeds into FBL[E]. However, some traces of this structure can still be recovered in some situations. To clarify this, let X be a Banach lattice, and consider the identity map idX : X → X that can be “extended” to a lattice homomorphism from FBL[X] to X which, in analogy with the Lipschitz-free space situation, will be denoted by βX . That is, βX : FBL[X] → X is the unique lattice homomorphism such that βX ◦ δX = idX , and we have kβX k = 1. Our purpose in this paper is to explore under which conditions there exists a lattice homomorphism T : X → FBL[X] such that βX ◦ T = idX . Note that, if this is the case, then X is lattice isomorphic (via T ) to a sublattice of FBL[X] which is lattice-complemented (via βX ).

Definition 1.1. We say that a Banach lattice X has the lattice-lifting property if there exists a lattice homomorphism T : X → FBL[X] such that βX ◦ T = idX . If in addition T can be chosen with kT k = 1, then we say that X has the isometric lattice-lifting property.

The paper is organized as follows. In Section 2, we present some basic properties and examples of Banach lattices with the lattice-lifting property: these include projective Banach lattices and the space FBL[E] for any Banach space E. As a consequence of the latter, a Banach lattice X has the lattice-lifting property if and only if it is lattice isomorphic to a lattice-complemented sublattice of FBL[X] (Corollary 2.8). In the first part of Section 3 we discuss the lattice-lifting property for the Banach lattice C(K), where K is a compact Hausdorff topological space. It turns out that the lattice-lifting property of C(K) implies ∗ that K is a neighborhood retract of the closed dual unit ball BC(K)∗ (equipped with the w -topology), see Theorem 3.2. This, combined with a result of [2], ensures that, for metrizable K, the lattice-lifting property of C(K) is equivalent to being a projective Banach lattice (Corollary 3.3). In the second part of Section 3 we analyze some topological properties of the set of all real-valued lattice homomorphisms on a Banach lattice having the isometric lattice-lifting property. Finally, in Section 4 we show that every Banach space with a 1-unconditional basis (as a Banach lattice with the coordinatewise order) satisfies the isometric lattice-lifting property (Theorem 4.1). This is the most technical result of the paper and its proof is based on some ideas of [3], where the particular case of c0 was addressed. A GODEFROY-KALTON PRINCIPLE FOR FREE BANACH LATTICES 3

Terminology. All our Banach spaces are real. Given a Banach space E, its norm is denoted by k·kE or ∗ simply k·k. We write BE := {x ∈ E : kxk≤ 1} for the closed unit ball of E and the symbol E stands for the topological dual of E. The weak∗-topology on E is denoted by w∗. By a lattice homomorphism between Banach lattices we mean a bounded linear operator preserving lattice operations. Given a Banach lattice X, we write X+ := {x ∈ X : x ≥ 0}. By a “sublattice” of X we mean a closed vector sublattice. We say that a sublattice Y ⊆ X is lattice-complemented if there is ∗ a lattice homomorphism P : X → Y such that P |Y = idY . We denote by Hom(X, R) ⊆ X the set of all lattice homomorphisms from X to R. In the case of free Banach lattices over Banach spaces this set admits a concrete description. Namely, given a Banach space E, the set Hom(FBL[E], R) is precisely ∗ ∗ {ϕx∗ : x ∈ E }, where each functional ϕx∗ : FBL[E] → R is defined by ∗ ϕx∗ (f) := f(x ) for every f ∈ FBL[E], see [5, Corollary 2.7]. By a “compact space” we mean a compact Hausdorff topological space. Given a compact space K, we denote by C(K) the Banach lattice of all real-valued continuous functions on K, equipped with the supremum norm and the pointwise order. Recall that a Schauder basis (en)n∈N of a Banach space E is called a 1-unconditional basis if for every n ∈ N, and a1,...,an,b1,...,bn ∈ R satisfying |ai| ≤ |bi| for i =1,...,n, then n n aiei ≤ biei .

i=1 i=1 X X In this case, E becomes a Banach lattice when equipped with the coordinatewise order given by

∗ ∗ N x ≤ y ⇐⇒ en(x) ≤ en(y) for all n ∈ , ∗ ∗ where (en)n∈N is the sequence in E of biorthogonal functionals associated with (en)n∈N.

2. Basics on the lattice-lifting property The following proposition shows some statements which are easily seen to be equivalent to the lattice- lifting property: Proposition 2.1. Let X be a Banach lattice and λ ≥ 1. The following statements are equivalent:

(i) There is a lattice homomorphism α : X → FBL[X] with kαk≤ λ such that βX ◦ α = idX . (ii) For every commutative diagram π Z / Z 1 ❆ 2 ` ❆❆ O ❆❆ T ❆❆ φ ❆❆ ❆❆ X

where Z1 and Z2 are Banach lattices, π and φ are lattice homomorphisms and T is a bounded linear ′ ′ ′ operator, there is also a lattice homomorphism T : X → Z1 with π ◦ T = φ and kT k≤ λkT k. (iii) If Y is a Banach lattice, π : Y → X is a lattice homomorphism and there is a bounded linear ′ operator T : X → Y such that π ◦ T = idX , then there is also a lattice homomorphism T : X → Y ′ ′ such that π ◦ T = idX and kT k ≤ λkT k.

Proof. (i)⇒(ii) Let Tˆ : FBL[X] → Z1 be the unique lattice homomorphism satisfying Tˆ ◦ δX = T . Since π ◦ Tˆ and φ ◦ βX are lattice homomorphisms and π ◦ Tˆ ◦ δX = φ = φ ◦ βX ◦ δX , we have π ◦ Tˆ = φ ◦ βX and so T ′ := Tˆ ◦ α satisfies the required properties. (ii)⇒(iii) Observe that (iii) is the particular case of (ii) when Z1 = Y , Z2 = X and φ = idX . (iii)⇒(i) Observe that (i) is the particular case of (iii) when Y = FBL[X], π = βX and T = δX .  4 A.AVILES,´ G. MART´INEZ-CERVANTES, J. RODR´IGUEZ, AND P. TRADACETE

We say that a Banach lattice X has the lattice-lifting property with constant λ if X and λ satisfy any of the equivalent conditions of Proposition 2.1. Projective Banach lattices provide a source of examples satisfying the lattice-lifting property, see Propo- sition 2.2 below. Given λ ≥ 1, a Banach lattice X is called λ-projective if whenever Y is a Banach lattice, J is a closed ideal of Y and Q : Y → Y/J is the quotient map, then for every lattice homomorphism T : X → Y/J there is a lattice homomorphism Tˆ : X → Y such that T = Q ◦ Tˆ and kTˆk ≤ λkT k. A Banach lattice is called projective if it is λ-projective for some λ ≥ 1. This class of Banach lattices has been studied in [2, 3, 7]. Examples of projective Banach lattices include all finite dimensional Banach lattices, all free Banach lattices of the form FBL(A) and the spaces ℓ1 and C[0, 1], see [7, §10 and §11]. Proposition 2.2. Every λ-projective Banach lattice has the lattice-lifting property with constant λ.

Proof. Suppose X is a λ-projective Banach lattice for some constant λ ≥ 1. Since βX : FBL[X] → X is a surjective lattice homomorphism, it factors as βX = S ◦ Q, where

Q : FBL[X] → FBL[X]/ ker(βX ) is the canonical quotient map and S : FBL[X]/ ker(βX ) → X −1 is a lattice isometry. Now, we consider the lattice isometry T := S : X → FBL[X]/ ker(βX ). By hypothesis, there exists a lattice homomorphism α : X → FBL[X] such that T = Q ◦ α and kαk ≤ λ. Therefore, we have βX ◦ α = S ◦ Q ◦ α = S ◦ T = idX , as required.  Projectivity for free Banach lattices is a quite restrictive property, namely: if E is a Banach space without the Schur property, then FBL[E] is not projective, see [2, Theorem 1.3]. However, all free Banach lattices satisfy the isometric lattice-lifting property (this is entirely analogous to [9, Lemma 2.10]): Proposition 2.3. FBL[E] has the isometric lattice-lifting property for every Banach space E. Proof. Consider the isometric embedding

T := δFBL[E] ◦ δE : E → FBL[FBL[E]] and the lattice homomorphism Tˆ : FBL[E] → FBL[FBL[E]] satisfying Tˆ ◦ δE = T and kTˆk = kT k = 1. ˆ We claim that βFBL[E] ◦ T = idFBL[E]. Indeed, given any x ∈ E we have ˆ (βFBL[E] ◦ T )(δE(x)) = βFBL[E](T (x)) = βFBL[E] δFBL[E](δE (x)) = δE(x). ˆ Since βFBL[E] ◦T and idFBL[E] are lattice homomorphisms and the sublattice generated
by {δE(x): x ∈ E} ˆ  is FBL[E], it follows that βFBL[E] ◦ T = idFBL[E]. Any Banach lattice X having the lattice-lifting property is lattice isomorphic to a sublattice of FBL[X]. Therefore, any property of free Banach lattices which is inherited by sublattices holds for Banach lattices having the lattice-lifting property. Corollary 2.5 below is obtained via this argument when applied to a couple of Banach lattice properties. A Banach lattice X is said to satisfy the σ-bounded chain condition if X+ \{0} can be written as a countable union X+ \{0} = n≥2 Fn such that, for each n ≥ 2 and each G⊆Fn with cardinality n, there exist distinct x, y ∈ G such that x ∧ y 6= 0. The σ-bounded chain condition is stronger than the countable S chain condition (every uncountable subset of X+ contains two distinct elements which are not disjoint).

Remark 2.4. Let X be a Banach lattice for which there is a sequence (φn) in Hom(X, R) which separates the points of X. Then X has the σ-bounded chain condition. Indeed, it suffices to consider

Fn := {x ∈ X+ : φm(x) 6= 0 for some m

Corollary 2.5. If a Banach lattice X has the lattice-lifting property, then: (i) Hom(X, R) separates the points of X. (ii) X satisfies the σ-bounded chain condition.

Proof. For any Banach space E, the set Hom(FBL[E], R) separates the points of FBL[E] (bear in mind the description of Hom(FBL[E], R) given in the introduction) and FBL[E] satisfies the σ-bounded chain condition, see [4, Theorem 1.2]. Both properties are clearly inherited by sublattices. 

Example 2.6. (i) Lp(µ) fails the lattice-lifting property for any 1 ≤ p ≤ ∞ whenever the measure µ is not purely atomic, because in this case Hom(Lp(µ), R) does not separate those functions supported on the non-atomic part of the measure space. (ii) c0(Γ) and ℓp(Γ) fail the lattice-lifting property for any 1 ≤ p ≤ ∞ whenever Γ is an uncountable set, because such Banach lattices fail the countable chain condition.

In Example 3.8 and Corollary 3.9 below we will show that, in general, the lattice-lifting property is not inherited by sublattices. However, for lattice-complemented sublattices the situation is different:

Proposition 2.7. Let X be a Banach lattice having the lattice-lifting property. If Y ⊆ X is a lattice- complemented sublattice, then Y has the lattice-lifting property.

Proof. Let j : Y → X denote the identity embedding and let P : X → Y be a lattice homomorphism such that P |Y = idY . Let αX : X → FBL[X] be a lattice homomorphism such that βX ◦ αX = idX . Let S : FBL[X] → FBL[Y ] be the unique lattice homomorphism satisfying S ◦ δX = δY ◦ P . We claim that the lattice homomorphism αY := S ◦ αX ◦ j : Y → FBL[Y ] satisfies βY ◦ αY = idY . Indeed, since both βY ◦ S and P ◦ βX are lattice homomorphisms and

βY ◦ S ◦ δX = βY ◦ δY ◦ P = P = P ◦ βX ◦ δX , we have βY ◦ S = P ◦ βX . Hence

βY ◦ αY = βY ◦ S ◦ αX ◦ j = P ◦ βX ◦ αX ◦ j = idY , as claimed.  As a consequence of Propositions 2.3 and 2.7 we have:

Corollary 2.8. A Banach lattice X has the lattice-lifting property if and only if it is lattice isomorphic to a lattice-complemented sublattice of FBL[X].

Remark 2.9. In general, the lattice-lifting property is not inherited by lattice quotients. For instance, the space L1[0, 1] fails the lattice-lifting property (see Example 2.6) and is a lattice quotient of FBL[ℓ1] (which has the lattice-lifting property, by Proposition 2.3). Indeed, since L1[0, 1] is a separable Banach space, there is a surjective bounded linear operator T : ℓ1 → L1[0, 1], hence the unique lattice homomorphism ˆ ˆ T : FBL[ℓ1] → L1[0, 1] satisfying T ◦ δℓ1 = T is surjective as well.

3. Retracts and the lattice-lifting property Given a topological space T , a subspace A ⊆ T said to be a retract (resp., neighborhood retract) of T if there is a continuous map from T to A which is the identity on A (resp., there is an open set U ⊆ T containing A such that A is a retract of U). Such a map is called a retraction. Absolute neighborhood retracts play an important role in the study of projective Banach lattices (see [7] and [2]). They also appear in a natural way when dealing with the lattice-lifting property. Recall that a compact space is said to be an absolute retract or AR (resp., absolute neighborhood retract or ANR) if it is a retract (resp., neighborhood retract) of any compact space containing it. The following elementary characterization will be useful: 6 A.AVILES,´ G. MART´INEZ-CERVANTES, J. RODR´IGUEZ, AND P. TRADACETE

Fact 3.1. A compact space K is an AR (resp. an ANR) if and only if it is homeomorphic to a retract (resp., neighborhood retract) of [0, 1]Γ for some non-empty set Γ.

The main result of this section is:

Theorem 3.2. Let K be a compact space such that C(K) has the lattice-lifting property. Then K is a ∗ neighborhood retract of (BC(K)∗ , w ). If in addition K is metrizable, then K is an ANR.

∗ As usual, we identify K with its canonical homeomorphic copy inside BC(K)∗ (with the w -topology), which consists of all evaluation functionals of the form h 7→ h(t) for every h ∈ C(K), where t ∈ K. We stress that, for an arbitrary compact space K, the space C(K) is projective whenever K is an ANR, see [2, Theorem 1.4]. So, bearing in mind Proposition 2.2 and Theorem 3.2, we can summarize the case of metrizable compacta as follows.

Corollary 3.3. Let K be a compact metrizable space. The following statements are equivalent: (i) K is an ANR. (ii) C(K) is projective. (iii) C(K) has the lattice-lifting property.

In order to prove Theorem 3.2 we need the following proposition:

Proposition 3.4. Let K and L be compact spaces such that K ⊆ L. Suppose that:

(i) There is a bounded linear operator E : C(K) → C(L) such that E(f)|K = f for all f ∈ C(K). (ii) C(K) has the lattice-lifting property. Then K is a neighborhood retract of L.

Proof. Let R : C(L) → C(K) be the surjective lattice homomorphism given by R(g) := g|K for all g ∈ C(L). Since R◦E = idC(K) and C(K) has the lattice-lifting property, there is a lattice homomorphism S : C(K) → C(L) such that R ◦ S = idC(K) (Proposition 2.1). By the general representation of lattice homomorphisms between spaces of continuous functions on compact spaces (see, e.g., [1, Theorem 2.34]), there exist a continuous map w : L → [0, ∞) and a map π : L → K which is continuous on L \ w−1({0}) such that S is given by S(f)(y)= w(y)f(π(y)) for all y ∈ L and all f ∈ C(K).

The fact that R ◦ S = idC(K) means that w(x)f(π(x)) = f(x) for all x ∈ K and all f ∈ C(K). Clearly, this implies that π is the identity on K and that w(x) = 1 for all x ∈ K. Thus, K is a retract of the open set L \ w−1({0}). 

∗ Proof of Theorem 3.2. Write L := BC(K)∗ (equipped with the w -topology) and consider the bounded linear operator E : C(K) → C(L) given by

E(f)(µ) := f dµ for all f ∈ C(K) and all µ ∈ L. ZK Here we identify the elements of C(K)∗ with regular signed measures on the Borel σ-algebra of K, via Riesz’s representation theorem. Since E(f)|K = f for every f ∈ K and C(K) has the lattice-lifting property, we can apply Proposition 3.4 to deduce that K is a neighborhood retract of L. Suppose now that K is metrizable. If K is finite, then it is easy to check that it is an AR. If K is infinite, then L is an infinite-dimensional compact convex metrizable subset of the locally convex space (C(K)∗, w∗) and so L is homeomorphic to the Hilbert cube [0, 1]N by Keller’s theorem (see, e.g., [8, Theorem 12.37]). Bearing in mind Fact 3.1, we conclude that K is an ANR.  A GODEFROY-KALTON PRINCIPLE FOR FREE BANACH LATTICES 7

We do not know whether the metrizability assumption can be removed from the second statement of Theorem 3.2:

Problem 3.5. Let K be a compact space such that C(K) has the lattice-lifting property. Is K an ANR?

Remark 3.6. If K is a compact space such that C(K) has the lattice-lifting property, then K is a finite union of pathwise connected closed subsets.

∗ Proof. As before, we consider BC(K)∗ equipped with the w -topology. By Theorem 3.2, there exist an open set K ⊆ V ⊆ BC(K)∗ and a continuous function r : V → K such that r|K = idK . For each x ∈ K we choose a convex open set Vx ⊆ BC(K)∗ such that x ∈ Vx ⊆ Vx ⊆ V . Since K is compact, there exist n n finitely many x1,...,xn ∈ K such that K ⊆ i=1 Vxi . Then K = i=1 r(Vxi ). Each Vxi is compact and pathwise connected (in fact, it is convex) and so r(V ) is compact and pathwise connected as well.  S xi S Example 3.7. Let c be the Banach lattice of all convergent sequences of real numbers (with the coordi- natewise order). (i) c fails the lattice-lifting property. Indeed, c is lattice isometric to C(N ∪ {∞}), where N ∪ {∞} denotes the one-point compactification of N equipped with the discrete topology. The only con- nected non-empty subsets of N ∪ {∞} are the singletons, so Remark 3.6 implies that c fails the lattice-lifting property. (ii) For each n ∈ N, let φn ∈ Hom(c, R) be the nth-coordinate projection. Since the sequence (φn) sep- arates the points of c, this space satisfies the σ-bounded chain condition (Remark 2.4). Therefore, the converse of Corollary 2.5 does not hold in general. (iii) We have a short exact sequence of lattice homomorphisms

j L 0 −→ c0 −→ c −→ R −→ 0,

where j is the canonical embedding and L is the map that takes each sequence to its limit. On the one hand, R is 1-projective and so it has the isometric lattice-lifting property. On the other hand, c0 has the isometric lattice-lifting property as a consequence of Theorem 4.1 below (cf. [3]). This shows that the isometric lattice-lifting property is not a 3-space property.

The following example and corollary show, in particular, that the lattice-lifting property is not stable under taking sublattices.

Example 3.8. Let T ⊆ R2 be the unit circumference. Then C(T) has the lattice-lifting property and contains a sublattice failing the lattice-lifting property. Indeed, the first statement follows from Corollary 3.3 because T is an ANR, since it is a neighborhood retract of [−1, 1]2 (recall Fact 3.1). Let K := TN (with the product topology). Observe that K is metrizable but it is not an ANR. Indeed, if it were, as an infinite countable product of non-empty separable metric spaces, all its factors but finitely many would be AR (see, e.g., [13, proof of Theorem 1.5.8]). However, T is not an AR since it is not a retract of [−1, 1]2 (see, e.g., [13, Theorem 3.5.5]). From Theorem 3.2 it follows that C(K) fails the lattice-lifting property. We will show that C(K) is lattice isometric to a sublattice of C(T). To this end, it suffices to check that there is a continuous surjection f : T → K, because in this case the composition map Tf : C(K) → C(T) given by Tf (g) := g ◦ f for all g ∈ C(K) would be an isometric lattice embedding. Now, the existence of such an f follows from the fact that [0, 1]N is a continuous image of [0, 1], by the Hahn-Mazurkiewicz theorem (see, e.g., [14, Theorem 31.5]), bearing in mind that T is a continuous image of [0, 1] and vice versa.

Corollary 3.9. Let E be a Banach space with dim(E) ≥ 2. Then FBL[E] contains a sublattice failing the lattice-lifting property. 8 A.AVILES,´ G. MART´INEZ-CERVANTES, J. RODR´IGUEZ, AND P. TRADACETE

Proof. Let E0 ⊆ E be a two-dimensional subspace. Since E0 is complemented in E, we know that FBL[E0] is lattice isomorphic to a sublattice of FBL[E] (see [5, proof of Corollary 2.8]). But FBL[E0] is lattice isomorphic to C(T), see [7, Proposition 5.3]. The conclusion follows from Example 3.8. 

For any Banach lattice X, the set Hom(X, R) is w∗-closed in X∗. Throughout the rest of this section ∗ we discuss some topological properties of the w -compact set Hom(X, R) ∩ BX∗ when X has the isometric lifting-property. We first introduce further terminology concerning homomorphisms on free Banach lattices. As we already mentioned, given a Banach space E, the set Hom(FBL[E], R) consists of all functionals ∗ ∗ ∗ ∗ ϕx∗ : FBL[E] → R defined by ϕx∗ (f) := f(x ) for every f ∈ FBL[E], where x ∈ E . For any set A ⊆ E we write ∗ ΦA := {ϕx∗ : x ∈ A}⊆ Hom(FBL[E], R).

∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ Clearly, the adjoint δE : FBL[E] → E satisfies δE(ϕx ) = x for every x ∈ E and its restriction ∗ R ∗ ∗ ∗ δE|Hom(FBL[E],R) : Hom(FBL[E], ) → E is a w -to-w continuous bijection. In particular, we have:

∗ ∗ ∗ Remark 3.10. Let E be a Banach space and K ⊆ E be a w -compact set. Then the map x 7→ ϕx∗ ∗ ∗ defines a homeomorphism between (K, w ) and (ΦK , w ).

∗ Proposition 3.11. Let X be a Banach lattice having the lattice-lifting property. Then (ΦHom(X,R), w ) is a retract of (Hom(FBL[X], R), w∗). If in addition X has the isometric lattice-lifting property, then ∗ ∗ (Hom(X, R) ∩ BX∗ , w ) is a retract of (BX∗ , w ).

Proof. Let T : X → FBL[X] be a lattice homomorphism such that βX ◦ T = idX . Note that for each x∗ ∈ Hom(X, R) we have ∗ (3.1) x ◦ βX = ϕx∗ ∗ ∗ (both are lattice homomorphisms and x ◦ βX ◦ δX = x = ϕx∗ ◦ δX ) and therefore ∗ ∗ ∗ ∗ ∗ ∗ ∗ (3.2) (βX ◦ T )(ϕx∗ )= βX (ϕx∗ ◦ T )= βX (x ◦ βX ◦ T )= βX (x )= ϕx∗ .

∗ ∗ ∗ ∗ ∗ ∗ ∗ R For any y ∈ X we have (βX ◦ T )(ϕy ) ∈ ΦHom(X,R), because T (ϕy ) ∈ Hom(X, ) and (3.1) yields

∗ ∗ ∗ ∗ (3.3) (βX ◦ T )(ϕy )= ϕT (ϕy∗ ). ∗ ∗ R ∗ ∗ So, the restriction of βX ◦ T is a retraction from (Hom(FBL[X], ), w ) onto (ΦHom(X,R), w ). Suppose now that X has the isometric lattice-lifting property, so that T can be chosen in such a way ∗ ∗ that kT k = 1. For each y ∈ BX∗ we have ϕy∗ ∈ BFBL[X]∗ , hence T (ϕy∗ ) ∈ Hom(X, R) ∩ BX∗ and ∗ ∗ ∗ ∗ ∗ so (βX ◦ T )(ϕy ) belongs to ΦHom(X,R)∩BX∗ (by (3.3)). Thus, the restriction of βX ◦ T is a retraction ∗ ∗ ∗ ∗ R ∗ from (ΦBX , w ) onto (ΦHom(X,R)∩BX∗ , w ). From Remark 3.10 it follows that (Hom(X, ) ∩ BX , w ) is ∗ a retract of (BX∗ , w ). 

A topological space T is said to be locally pathwise connected if for each t ∈ T and each open set U containing t there is an open set V containing t such that for every t′ ∈ V there is a continuous map h : [0, 1] → U such that h(0) = t and h(1) = t′. This property is inherited by retracts (see, e.g., [10, Proposition 10.2]) and holds for any convex subset of a locally convex space. So, from the previous proposition we get:

∗ Corollary 3.12. Let X be a Banach lattice. Then (Hom(X, R) ∩ BX∗ , w ) is locally pathwise connected whenever X has the isometric lattice-lifting property.

Another application of Proposition 3.11 is the following:

∗ Corollary 3.13. Let X be a separable Banach lattice. Then (Hom(X, R) ∩ BX∗ , w ) is an AR whenever X has the isometric lattice-lifting property. A GODEFROY-KALTON PRINCIPLE FOR FREE BANACH LATTICES 9

∗ n Proof. Note that (BX∗ , w ) is homeomorphic to either [0, 1] for some n ∈ N (if X is finite-dimensional) or [0, 1]N (if X is infinite-dimensional) by Keller’s theorem (see, e.g., [8, Theorem 12.37]). By Fact 3.1 and ∗ Proposition 3.11, (Hom(X, R) ∩ BX∗ , w ) is an AR.  We wonder whether the separability assumption can be removed from the previous statement:

∗ Problem 3.14. Let X be a Banach lattice. Is (Hom(X, R) ∩ BX∗ , w ) an AR whenever X has the isometric lattice-lifting property?

4. Banach spaces with a 1-unconditional basis A Banach space with a 1-unconditional basis becomes naturally a Banach lattice when equipped with the coordinatewise order. The purpose of this section is to prove the following: Theorem 4.1. Let X be a Banach space with a 1-unconditional basis. If we consider X as a Banach lattice with the coordinatewise order, then X has the isometric lattice-lifting property. The proof of Theorem 4.1 requires some auxiliary lemmata. Recall that the free Banach lattice FBL(A) over a non-empty set A can be identified with FBL[ℓ1(A)], see [5, Corollary 2.9]. For each a ∈ A, we ′ ′ denote by ea ∈ ℓ1(A) the vector defined by ea(a ) = 0 for all a ∈ A \{a} and ea(a) = 1, and we write

δa := δℓ1(A)(ea) ∈ FBL(A) for simplicity. Lemma 4.2. Let E be a Banach space. The vector lattice homomorphism ∗ RE ∗ ∗ ϕ : FBL(BE) → , ϕ(f)(x ) := f(x |BE ), satisfies the following properties:

(i) ϕ(FBL(BE)) ⊆ FBL[E]. (ii) ϕ has norm 1 and dense range as a bounded linear operator from FBL(BE) to FBL[E].

∗ ∗ ∗ Proof. Fix f ∈ FBL(BE). Clearly, ϕ(f) is positively homogeneous. Given finitely many x1,...,xn ∈ E n ∗ with supx∈BE k=1 |xk(x)|≤ 1, we have n n P ∗ ∗ |ϕ(f)(xk)| = |f(xk|BE )|≤kfkFBL(BE), k k X=1 X=1 so ϕ(f) ∈ H0[E] and kϕ(f)kFBL[E] ≤kfkFBL(BE). Therefore, ϕ is a lattice homomorphism from FBL(BE) to H0[E] with kϕk≤ 1. Let S be the vector sublattice of FBL(BE) generated by {δx : x ∈ BE}. Since ϕ(δx)= δE(x) for every x ∈ BE (which implies that kϕk = 1) and ϕ is a lattice homomorphism, ϕ(S) is the vector sublattice of H0[E] generated by {δE(x): x ∈ BE}. Hence ϕ(S) is a dense vector sublattice of FBL[E]. Since S is dense in FBL(BE) and ϕ is continuous, we conclude that ϕ(FBL(BE )) is a dense vector sublattice of FBL[E].  Definition 4.3. Let E be a Banach space, B ⊆ E be a non-empty set and f : E∗ → R be a function. We ∗ ∗ ∗ ∗ ∗ ∗ ∗ say that f depends on coordinates of B if f(x )= f(y ) for every x ,y ∈ E such that x |B = y |B. Lemma 4.4. Let E be a Banach space and f : E∗ → R be a positively homogeneous function such that:

(i) f|BE∗ is norm-continuous; (ii) there is a finite non-empty set B ⊆ E such that f depends on coordinates of B. Then f ∈ FBL[E]. Lemma 4.4 will be obtained as a consequence of a particular case via Lemma 4.2. We isolate such particular case in Lemma 4.5 below. It is essentially based on the description of the space FBL(A) for a finite set A given in [7, Proposition 5.3], but we include a detailed proof for the reader’s convenience. 10 A. AVILES,´ G. MART´INEZ-CERVANTES, J. RODR´IGUEZ, AND P. TRADACETE

Lemma 4.5. Let A be a non-empty set and f : ℓ∞(A) → R be a positively homogeneous function such that: ∗ (i) f|Bℓ∞(A) is w -continuous; (ii) there is a finite non-empty set A0 ⊆ A such that f depends on coordinates of {ea : a ∈ A0}. Then f ∈ FBL(A).

Proof. Let r : ℓ∞(A) → ℓ∞(A) be the bounded linear operator given by

∗ ∗ ∗ r(x ) := x χA0 for all x ∈ ℓ∞(A).

Note that a function g : ℓ∞(A) → R depends on coordinates of {ea : a ∈ A0} if and only if g = g ◦ r. Observe also that ∗ ∗ S := {x ∈ r(ℓ∞(A)) : kx kℓ∞(A) =1} ∗ is a w -compact subset of ℓ∞(A) (because A0 is finite). Claim 1. If g : ℓ∞(A) → R is a positively homogeneous function which depends on coordinates ∗ of {ea : a ∈ A0}, then kgkFBL[E] ≤ supx∗∈S |g(x )| · |A0|. In order to prove this, we suppose that ∗ ∗ ∗ C := supx∗∈S |g(x )| < ∞ (otherwise the inequality is trivial). Take x1,...,xk ∈ ℓ∞(A) such that k ∗ ∗ ∗ supa∈A i=1 |xi (a)|≤ 1. For each i ∈{1,...,k} we choose ai ∈ A0 such that |xi (ai)| = kr(xi )kℓ∞(A) and we take y∗ ∈ S such that r(x∗)= |x∗(a )| y∗. Since g = g ◦ r and is positively homogeneous, we have Pi i i i i k k k k ∗ ∗ ∗ ∗ ∗ |g(xi )| = |g(r(xi ))| = |g(yi )| · |xi (ai)|≤ C |xi (ai)| i=1 i=1 i=1 i=1 X X X X k k ∗ ∗ = C |xi (a)|≤ C |xi (a)|≤ C · |A0|.

a∈A0 i=1 a∈A0 i=1 X aXi=a X X

This shows that kgkFBL[E] ≤ C · |A0| and finishes the proof of Claim 1. ∗ Recall that each g ∈ FBL(A) is w -continuous on bounded subsets of ℓ∞(A), see [5, Lemma 4.10]. Let j : FBL(A) → C(S) be the restriction map, that is,

j(g) := g|S for all g ∈ FBL(A). Then j is a lattice homomorphism with kjk≤ 1. Let V be the set of all f ∈ FBL(A) depending on coordinates of {ea : a ∈ A0}. Clearly, V is a sublattice of FBL(A) and δa ∈ V for all a ∈ A0. Therefore, the sublattice U generated by {δa : a ∈ A0} is contained in V . Claim 2. The equality j(U) = C(S) holds. Indeed, since j is a lattice homomorphism, j(U) is a vector sublattice of C(S). By the inclusion U ⊆ V and Claim 1, j(U) is norm-closed and j|U is an isomorphism between U and j(U) as Banach spaces. So, by the Stone-Weierstrass theorem, in order to prove that j(U) = C(S) it suffices to check that j(U) separates the points of S and that the constant function 1 ∈ C(S) belongs to j(U). On the one hand, if x∗,y∗ ∈ S are distinct, then x∗(a) 6= y∗(a) for ∗ ∗ some a ∈ A0, that is, j(δa)(x ) 6= j(δa)(y ). Hence j(U) separates the points of S. On the other hand, let ϕ := |δ | ∈ U. Then j(ϕ)= |j(δ )| and so for every x∗ ∈ S we have a∈A0 a a∈A0 a ∗ ∗ ∗ W j(ϕ)(xW ) = max |j(δa)|(x ) = max |x (a)| =1, a∈A0 a∈A0 that is, j(ϕ) is the constant function 1 ∈ C(S). The proof of Claim 2 is finished. Finally, let f : ℓ∞(A) → R be as in the statement of the lemma. By Claim 2, there is g ∈ U such that f|S = g|S. Since f and g are positively homogeneous and depend on coordinates of {ea : a ∈ A0}, we have f = g, as it can be easily checked. Thus f ∈ FBL(A).  A GODEFROY-KALTON PRINCIPLE FOR FREE BANACH LATTICES 11

Proof of Lemma 4.4. Write B = {x1,...,xn}. We can assume without loss of generality that B ⊆ BE and ∗ ∗ ∗ that B is linearly independent. The latter condition ensures that we can find x1,...,xn ∈ E such that ∗ ∗ R xi (xj ) = 1 if i = j and xi (xj ) = 0 if i 6= j for every i, j ∈{1,...,n}. Define g : ℓ∞(BE ) → by

∗ ∗ g(ξ) := f ξ(x1)x1 + ··· + ξ(xn)xn for all ξ ∈ ℓ∞(BE ).

Clearly, g is positively homogeneous, depends on coordinates
of {ex ,...,ex } and the restriction g|B 1 n ℓ∞(BE ) ∗ is w -continuous (because f is norm-continuous on bounded sets), so g ∈ FBL(BE) by Lemma 4.5. Since f depends on coordinates of {x1,...,xn}, we have

∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ g(x |BE )= f x (x1)x1 + ··· + x (xn)xn = f(x ) for all x ∈ E , and so we can apply Lemma 4.2 to conclude that f ∈ FBL[
E]. 

∗ Lemma 4.6. Let E be a Banach space with a 1-unconditional basis (en)n∈N and let (en)n∈N be the sequence ∗ in E of biorthogonal functionals associated with (en)n∈N. Then

l l ∗ ∗ ∗ |xi (emi )| emi ≤ sup |xi (x)| ∗ x∈BE i=1 E i=1 X X N ∗ ∗ ∗ ∗ N for all l ∈ , all x1, x2,...,xl ∈ BE and all m1,m 2,...,ml ∈ .

Proof. Consider m := max{mi : i =1,...,l}. Fix z ∈ BE. We will show that

l l ∗ ∗ ∗ |xi (emi )| emi (z) ≤ sup |xi (x)| =: α. x∈BE i=1 i=1 X X m m ε m ∗ ε For every choice of signs ε = ( εj )j=1 ∈ {−1, 1} , define z := j=1 εjej (z)ej , so that z ∈ BE (since the basis is 1-unconditional) and therefore P l m l ε e∗(z)x∗(e ) = |x∗(zε)|≤ α. j j i j i i=1 j=1 i=1 X X X

So, in order to finish the proof it suffices to check that

l l m (4.1) |x∗(e )| e∗ (z) ≤ max ε e∗(z)x∗(e ) : ε ∈ {−1, 1}m . i mi mi  j j i j  i=1 i=1 j=1 X X X 

To this end, observe first that   l m l m 1 (4.2) max ε e∗(z)x∗(e ) : ε ∈ {−1, 1}m ≥ ε e∗(z)x∗(e )  j j i j  2m j j i j i=1 j=1 ε∈{−1,1}m i=1 j=1 X X  X X X

  l m l m 1 1 = ε e∗(z)x∗(e ) + ε e∗(z)x∗(e ) 2m j j i j 2m j j i j i=1 ε∈{−1,1}m j=1 i=1 ε∈{−1,1}m j=1 X εX=1 X X ε X=−1 X mi mi

l m l m 1 ∗ ∗ 1 ∗ ∗ ≥ εj e (z)x (ej ) + εj e (z)x (ej ) . 2m j i 2m j i i=1 ε∈{−1,1}m j=1 i=1 ε∈{−1,1}m j=1 X εX=1 X X ε X=−1 X mi mi

12 A. AVILES,´ G. MART´INEZ-CERVANTES, J. RODR´IGUEZ, AND P. TRADACETE

Now, for each i =1 ...,l and each σ ∈ {−1, 1} we have

m ∗ ∗ εj e (z)x (ej ) j i ε∈{−1,1}m j=1 εmX=σ X i

m m−1 ∗ ∗ ∗ ∗ m−1 ∗ ∗ = σ2 e (z)x (em )+ εj e (z)x (ej ) =2 |e (z)x (em )|. mi i i j i mi i i ε∈{−1,1}m j=1 εmX=σ jX6=mi i

Thus, from (4.2) it follows that

l m l l max ε e∗(z)x∗(e ) : ε ∈ {−1, 1}m ≥ |e∗ (z)x∗(e )|≥ |x∗(e )| e∗ (z) ,  j j i j  mi i mi i mi mi i=1 j=1 i=1 i=1 X X  X X and so (4.1) holds. The proof is finished.    We are now ready to prove Theorem 4.1.

Proof of Theorem 4.1. We consider X equipped with the coordinatewise order induced by a fixed 1- unconditional basis (en)n∈N of X. Clearly, we can assume without loss of generality that kenkX = 1 for all n ∈ N. We will show that there is a disjoint sequence (fn)n∈N ∈ FBL[X]+ such that βX (fn)= en for all n ∈ N and N N anfn ≤ anen

n=1 FBL[X] n=1 X X X for every N ∈ N and all a1,...,aN ∈ R. Once we have this, it is easy to check that

∗ T : X → FBL[X],T (x) := en(x)fn, n N X∈ ∗ is a well-defined lattice homomorphism such that βX ◦T = idX and kT k = 1, where (en)n∈N is the sequence ∗ in X of biorthogonal functionals associated with (en)n∈N. We divide the proof into several steps.

N ∗ ∗ R Step 0. Identification of βX . For each n ∈ , both ϕen and en ◦ βX belong to Hom(FBL[X], ) and ∗ ∗ ∗ ∗ ∗ satisfy ϕen ◦ δX = en = en ◦ βX ◦ δX , hence ϕen = en ◦ βX . It follows that for every f ∈ FBL[X] we have

∗ (4.3) βX (f)= f(en)en, n N X∈ the series being unconditionally convergent.

Step 1. Construction of the fn’s. Let (Mn)n∈N and (Nn)n∈N be two strictly increasing sequences such that Mn

∗ ∗ + ∗ ∗ ∗ ∗ |x (en)|− Nn max {|x (em)| : mn gm |x (em)|/|x (en)| if x (en) 6= 0, fn(x ) := ∗ (0
Q
if x (en) = 0. A GODEFROY-KALTON PRINCIPLE FOR FREE BANACH LATTICES 13

∗ ∗ n+k ∗ ∗ Here m>n gm |x (em)|/|x (en)| is the limit of the sequence ( m=n+1 gm |x (em)|/|x (en)| k∈N, which is decreasing and contained in [0, 1] (because each g takes values in [0, 1]). Clearly, f is positively Q
m Q n
homogeneous. ∗ ∗ ∗ Step 2. (fn)n∈N is disjoint. Indeed, fix n Nl max {|x (em)| : m Nl|x (en)|. Hence gl(|x (el)|/|x (en)|) = 0 and so ∗ fn(x ) = 0. This shows that fn ∧ fl = 0. ∗ Step 3. fn ∈ FBL[X] for every n ∈ N. To prove this, fix n, k ∈ N and define hn,k : X → R by

∗ ∗ + n+k ∗ ∗ ∗ ∗ |x (en)|− Nn max {|x (em)| : m

We claim that hn,k ∈ FBL[X]. Indeed, observe that hn,k is positively homogeneous and depends on ∗ coordinates of the finite set {e1,...,en+k} ⊆ X. Further, it is easy to check that hn,k is w -continuous: ∗ ∗ ∗ bear in mind that the map x 7→ |x (em)| is w -continuous for every m ∈ N, that all the gm’s are continuous ∗ ∗ ∗ ∗ and that the inequality |hn,k(x )| ≤ |x (en)| holds for every x ∈ X . From Lemma 4.4 it follows that hn,k ∈ FBL[X], as claimed. We next prove that 1 (4.4) khn,k − fnkFBL[X] ≤ . Mj j>n k X+ ∗ ∗ ∗ l ∗ Indeed, take x1,...,xl ∈ X such that supx∈BX i=1 |xi (x)|≤ 1. We seek to estimate l P ∗ ∗ |hn,k(xi ) − fn(xi )|, i=1 X ∗ ∗ so we can assume that hn,k(xi ) 6= fn(xi ) for all i =1,...,l. This means that for each i =1,...,l we have ∗ ∗ ∗ ∗ ∗ |xi (en)| 6= 0andthereis mi >n+k such that gmi (|xi (emi )|/|xi (en)|) 6= 1, that is, |xi (emi )| >Mmi |xi (en)|. So, bearing in mind that hn,k ≥ fn, we have

∗ ∗ ∗ ∗ ∗ ∗ 1 ∗ |hn,k(xi ) − fn(xi )| = hn,k(xi ) − fn(xi ) ≤ hn,k(xi ) ≤ |xi (en)| < |xi (emi )| Mmi for every i =1,...,l. It follows that

l l l ∗ ∗ 1 ∗ ∗ ∗ 1 |hn,k(xi ) − fn(xi )| ≤ |xi (emi )| = |xi (emi )|emi ej Mm  Mj  i=1 i=1 i i=1 ! j>n k X X X X+ l ∞  ∞  ∗ ∗ 1 1 ≤ |xi (emi )|emi · ej ≤ , Mj Mj i=1 ∗ j>n+k j>n+k X X X X X where the last inequality is a consequence of Lemma 4.6. This works for any finite collection of functionals ∗ ∗ ∗ l ∗ x1,...,xl ∈ X satisfying supx∈BX i=1 |xi (x)|≤ 1, hence inequality (4.4) holds. Finally, since each hn,k belongs to FBL[X] and limk→∞ khn,k − fnkFBL[X] = 0 (by (4.4) and the fact 1 P that < ∞), we conclude that fn ∈ FBL[X], which finishes the proof of Step 3. j∈N Mj N StepP 4. βX (fn) = en for every n ∈ . Indeed, this follows from (4.3), bearing in mind that, for each N ∗ ∗ j ∈ , we have fn(ej ) = 1 if j = n and fn(ej ) = 0 if j 6= n (by the very definition of fn). Step 5. The inequality N N (4.5) anfn ≤ anen

n=1 FBL[X] n=1 X X X

14 A. AVILES,´ G. MART´INEZ-CERVANTES, J. RODR´IGUEZ, AND P. TRADACETE

holds for every N ∈ N and all a1,...,aN ∈ R. Indeed, since the sequence (fn)n∈N is disjoint and (en)n∈N is 1-unconditional, the norms appearing in (4.5) do not change by switching the signs of the coefficients, N ∗ ∗ ∗ ∗ and we can assume that an ≥ 0 for all n = 1,...,N. Set f := n=1 anfn. Fix x1, x2,...,xk ∈ X with sup k |x∗(x)|≤ 1, and suppose without loss of generality that f(x∗) 6= 0 for all j =1,...,k. Since x∈BX j=1 j P j the f ’s are disjoint, for each j =1,...,k there is a unique n ∈{1,...,N} such that f (x∗) 6= 0, hence n P j nj j k k k ∗ ∗ ∗ |f(xj )| = f(xj )= anj fnj (xj ). j=1 j=1 j=1 X X X ∗ ∗ Since fnj (xj ) ≤ |xj (enj )| for every j =1,...,k, we get

k k k N ∗ ∗ ∗ ∗ |f(x )|≤ an |x (en )| = |x (en )|e anen j j j j  j j nj  j=1 j=1 j=1 n=1 ! X X X X k  N  N ∗ ∗ ≤ |xj (enj )|enj · anen ≤ anen ,

j=1 ∗ n=1 X n=1 X X X X X where the last inequality follows from Lemma 4.6. This shows that (4.5) holds.

The proof of Theorem 4.1 is complete. 

Acknowledgements We wish to thank Valentin Ferenczi for suggesting this line of research. Research partially supported by Fundaci´on S´eneca [20797/PI/18] and Agencia Estatal de Investigaci´on [MTM2017-86182-P to A. Avil´es, G. Mart´ınez-Cervantes, and J. Rodr´ıguez; MTM2016-76808-P, MTM2016- 75196-P to P. Tradacete; all grants being cofunded by ERDF, EU]. The research of G. Mart´ınez-Cervantes was co-financed by the European Social Fund and the Youth European Initiative under Fundaci´on S´eneca [21319/PDGI/19]. P. Tradacete gratefully acknowledges support by Ministerio de Econom´ıa, Industria y Competitividad through “Severo Ochoa Programme for Centres of Excellence in R&D” [SEV-2015-0554] and Consejo Superior de Investigaciones Cient´ıficas through “Ayuda extraordinaria a Centros de Excelen- cia Severo Ochoa” [20205CEX001].

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Universidad de Murcia, Departamento de Matematicas,´ Facultad de Matematicas,´ 30100 Espinardo (Murcia), Spain. Email address: [email protected]

Universidad de Murcia, Departamento de Matematicas,´ Facultad de Matematicas,´ 30100 Espinardo (Murcia), Spain. Email address: [email protected]

Departamento de Ingenier´ıa y Tecnolog´ıa de Computadores, Facultad de Informatica,´ Universidad de Murcia, 30100 Espinardo (Murcia), Spain. Email address: [email protected]

Instituto de Ciencias Matematicas´ (CSIC-UAM-UC3M-UCM), Consejo Superior de Investigaciones Cient´ıficas, C/ Nicolas´ Cabrera, 13–15, Campus de Cantoblanco UAM, 28049 Madrid, Spain. Email address: [email protected]