Page 224 - Textos de Matemática Vol. 46

214 D. HOFMANN AND P. NORA
4. Generalised Esakia spaces as idempotent split completion
With the results of the last section in mind, we would like to conclude that GEsaDist is the idempotent split completion of the category CompHausRel. This follows indeed from Theorem 3.2, as soon as we know that the cate- gory StCompDist is idempotent split complete. Similarly to the case of spectral spaces, it is easier to argue in the dual category. We write StContDLatW,⌧ to denote the category having as objects continuous distributive lattices where the way-below relation is stable under finite infima, and as morphisms those maps which preserve suprema and the way-below relation. Note that every continuous distributive lattice is a frame. The following result can be found in [28].
Theorem 4.1. The category StCompDist is dually equivalent to the category StContDLatW,⌧ .
Proposition 4.2. The category StContDLatW,⌧ is idempotent split complete.
Proof. Let e: L ! L be an idempotent morphism in StContDLatW,⌧. Then e splits in the category of sup-lattices and sup-preserving maps, that is, there is a complete lattice M and sup-preserving maps r: L ! M and s: M ! L so that e = sr and rs = 1M . Then M is certainly a distributive lattice, and, since the embedding s : M ! L preserves suprema, for all x, y 2 M one has
s(x)⌧s(y) ) x⌧y.
Consequently, since e: L ! L preserves the way-below relation, so does r: L ! M. We show now that s: M ! L preserves the way-below relation. To this end, let x ⌧ y in M. Since L is a con_tinuous lattice,
s(y) = {b 2 L | b ⌧ s(y)},
and note that {b 2 L | b ⌧ s(y)} is directed. Hence, y = rs(y) is the directed supremum of {r(b) 2 L | b ⌧ s(y)}. Therefore there exist some b ⌧ s(y) with x  r(b) and, since e preserves the way-below relation, we obtain
s(x)  sr(b) = e(b) ⌧ e(s(y)) = s(y).
This shows that s preserves the way-below relation, and from that it follows that M is a continuous lattice. Finally, we prove that the way-below relation in M is stable under finite infima. Note that r(>) = > since r is surjective, therefore,since>⌧>inL,weobtain>⌧>inM.Letnowx⌧x0 and y⌧y0 inM.Then
x^y=r(s(x)^s(y))⌧r(s(x0)^s(y0))=x0 ^y0. ⇤