, vi) ? preserves interpolators: If a ? b, then ?(b|a) = ?(b)|?(a)

, Property (iv) above was considered in [27, p. 134], as part of the definition of callitic morphisms for distributive inverse semigroups. It is a strengthening of the condition that ? preserves ?, and is necessary since we consider non-Hausdorff inverse semigroupoids, It is easy to check that morphisms of ?-ordered inverse semigroups are stable under composition

, then by (iv) there exists c ? a ? b such that ?(u) ? ?(c) ? ?(a ? b), so ?(a ? b) is the largest ?-lower bound of {a, b}, i.e., ?(a ? b) = ?(a) ? ?(b). It is important to note that there are semigroup homomorphisms, between ?-ordered inverse semigroups, which satisfy Properties (i)-(v) but not (vi)

, and the compatible order x ? y ?? x = 0 or x = y, and let L 2 = {0, 1}, as an ideal of L 3 , and the restriction of ?. Both L 3 and L 2 are ?-ordered inverse semigroups. (L 3 is isomorphic to KB(L 2 ), where L 2 is seen simply as an inverse semigroup, and L 2 is isomorphic to KB({0}).) The map ?

, Similarly, the map ? : L 3 ? L 2 , ?(0) = ?(1) = 0, ?(2) = 1, satisfies all of (i)-(v) but not (vi)

, s(a) = s(b) and ?(a) = ?(b) implies a = b (i.e., ? is injective on all fibers), and star-surjective if for all t ? T and all a ? S, if s(t) = s(?(a)), then there exists b ? S with s(b) = s(a) and ?(b) = t. If ? is both star-injective and star-surjective, The category Amp ?. A homomorphism ? : S ? T of inverse semigroupoids is star-injective if for all a, b ? S

, On objects, to each ample inverse semigroup S, we set K(S) = (KB(S), ?)

, The map A ? ? ?1 (A) is a morphism of ?-ordered inverse semigroups. We thus define the functor K on a morphism ? : A ? B of Amp ? as K(?)(A), Lemma 5.16. Let ? : S ? T is a proper continuous covering homomorphism of ample semigroupoids. If A ? KB(T) then ? ?1 (A) ? KB(S)

, As ? is continuous and proper, then ? ?1 (A) is open and compact. Let us prove that it is a bisection of S. Suppose a, b ? ? ?1 (A) and s(a) = s(b)

, As ? is star-surjective, there exists p b ? S with s(p b ) = s(z) and ?(p b ) = b. Then s(?(pp * b )) = s(b * ) = s(a)

, As ? is star-injective then x = p a p b ? ? ?1 (A) ? ? ?1 (B). The first non-trivial property that we need to prove for K(?), to conclude that it is a morphism of ?-ordered inverse semigroups, is properness. Suppose K ? KB(S), Then s(p a p b ) = s(p b ) = s(z), and ?

, ? = S for every ample semigroupoid S. Given another ample semigroupoid T, a morphism ? : KB(S) ? KB(T)) thus induces a morphism P(?) : T ? S, and in particular a map between the underlying vertex spaces T (0) and S (0) , or alternatively by the content of Subsection 5, The functor P : ?-Ord op ? Amp. As a motivation, suppose that the functor P is defined in a manner that P(KB(S)), vol.1

Moreover, the map P(?) (0) thus defined is continuous and proper ,

, ) and s 1 ,. .. , s n ? E(S) such that t = n i=1 t i and t i ? ?(s i ) for each i. As F is ?-prime (Lemma 5.12), for some i we have t i ? F

, F) is upwards closed, does not contain 0, and contains ? ?1 (F), so to prove that it is a filter we need to prove that it is closed under products. For this

, We have f p(p \ q f ) ? q f (p \ q f ) = q f ? (p \ q f ) = 0, and because f ? P(?) (0) (F) then p(p \ q f ) ? ? ?1 (F), so (p \ q f ) ? ? ?1 (F) as well. We therefore have q f ? ? ?1 (F). Suppose now, in order to obtain a contradiction, that f g ? P(?) (0) (F), so that we may find e ? ? ?1 (F) such that f ge = 0, ) (F). Then f p ? p, so we may consider the interpolator q f. . = p|(f p)

, In fact, its definition makes it clear, by Lemma 5, vol.10

, ?-v), the join ?(e) ? a exists. As ? is proper, the same considerations as in the proof that ? ?1 (F) is nonempty allow us to assume that ?(e) ? a ? ?(g), and in particular ?(e) = ?(eg), particular, ?(e)a = 0 implies ?(p)a = 0, so ?(p) ? F. On the other hand, e(g \ p) = ge(g \ p) ? p(g \ p) = 0, and since e ? P(?) (0) (F) then g \ p ? ? ?1 (F), i.e., ?(g \ p) ? F, vol.5

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