A concrete co-existential map that is not confluent

In [11,1] Bankston gave an example of a co-existential map that is not confluent. The construction is rather involved and does not produce a concrete example of such a map. A lot of effort was needed to get the main ingredient, to wit a co-diagonal map that is not monotone.

The purpose of this note is to show that one can write down a concrete map between two rather simple continua that is co-existential and not confluent. It will be clear from the construction that the range space admits co-diagonal maps that are nor confluent and, a fortiori, not monotone.

In the interest of brevity we try to keep the notation down to the bare minimum. I (u) and we let q u = βπ Y ↾ Y u . In the terminology of [11,1] the space Y u is the ultra-copower of Y by the ultrafilter u and q u : Y u → Y is the associated co-diagonal map. A map f : X → Y between compact spaces is co-existential if there are a set I, an ultrafilter u on I and a map g : Y u → X such that q u = f • g.

These notions can be seen as dualizations of notions from model theory and they offer inroads to the study of compact Hausdorff spaces by algebraic, and in particular lattice-theoretic, means.

1.2. Two notions from continuum theory. On a first-order algebraic level there is not much difference between Y and Y u : they have elementarily equivalent latticebases for their closed sets: the map A → Y u ∩ cl β (A × I) is an elementary embedding of such bases. It is therefore not unreasonable to expect that the co-diagonal map q u be well-behaved. One could expect it, for example, to be confluent, which means that if C is a subcontinuum of Y then every component of q ← u [C] would be mapped onto C by q u . Certainly some component of q ← u [C] is mapped onto C: the component that contains Y u ∩ cl β (C × I) (this shows that q u is weakly confluent). Intuitively there should be no difference between the components, so all should be mapped onto C. The example below disproves this intuition.

The paper [11,1] gives (references for) other reasons why it is of interest to know whether co-diagonal and co-existential maps are confluent.

We start with the Closed Infinite Broom [33; 3, Example 120]

where H n = { t, t/2 n : 0 ≤ t ≤ 1} is the nth hair of the broom.

The range space is B with the limit hair extended to have length 2:

We denote the extended hair [0, 2] × {0} by H ω . The domain of the map is B with an extra hair of length 2 along the y-axis:

The map f : X → Y is the (more-or-less) obvious one:

Thus f is the identity on B and it rotates the points on the extra over -1 2 π. Claim 1. The map f is not confluent.

Proof. This is easy. The components of the preimage of the continuum C = [1, 2] × {0} are the interval {0} × [1,2] and the singleton { 1, 0 }; the latter does not map onto C. Claim 2. The map f is co-existential.

Proof. We need to find an ultrafilter u and a map g : Y u → X such that f • g is the co-diagonal map q u : Y u → Y . In fact any free ultrafilter u on ω will do.

We define two closed subsets F and G of Y × ω and define g on the intersections

It is an elementary verification that q u [F u ] = B and q u [G u ] = H ω . This allows us to define g : Y u → X by cases: on F u we define g to be just q u and on G u we define g = R • q u , where R rotates the plane over 1 2 π. These definitions agree at the point in F u ∩ G u and give continuous maps on F u and G u respectively. Therefore the combined map g : Y u → X is continuous as well.

This also shows that the co-diagonal map q u is not confluent: no component of the preimage under g of 1, 0 is mapped onto C.

Remark. In [22, 2] Bankston showed that if a continuum K is such that every coexistential map onto K is confluent then every K must be connected im kleinen at each of its cut points. The continuum Y above is connected im kleinen at all cut points but one: the point 1, 0 , so Y does not qualify as a counterexample to the converse.

To obtain a countereample multiply X and Y by the unit interval and multiply f by the identity. The proof that the new map is co-existential but not confluent is an easy adaptation of the proof that f has these properties. Since Y has no cut points it is connected im kleinen at all of them.

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