Re: Continuous maps between boolean/stone spaces

On 12.09.2007 10:07, Jannick Asmus wrote:
On 08.09.2007 20:35, Jannick Asmus wrote:
On 08.09.2007 00:10, Jose Capco wrote:
Dear NG,

I have this strange impression that continuous surjective maps between
boolean/stone spaces are also automatically open.. is this true?

I think this is true - up to checks of the sketch of proof given below:

I scanned the following which turned out to be wrong to find the spots where it breaks down.

By Stone's representation theorem (cf., e.g., http://planetmath.org/encyclopedia/MHStonesRepresentationTheorem.html) we can assume that the surjection of Stone spaces comes from a homomorphism phi: A -> B of Boolean algebras. We denote the corresponding continuous map of Stone spaces B* -> A* by phi*.

Here A* denotes the dual of A, i.e. A* = Hom(A,E) with the trivial Boolean algebra E with underlying set {0,1} (0 <> 1), A* is equipped with the topology induced by {0,1}^A which is a compact and Hausdorff space, by Tychonov. Since A* is closed in {0,1}^A, A* is compact and Hausdorff, too.

The assumption that phi* is surjective implies that phi is injective, hence we can assume w.l.o.g. that A is a subalgebra of B.

Let's assume in the first step that A and B are *free* Boolean algebras with basis S and T, resp., with S contained in T. Then the Stone spaces A* and B* are homeomorphic to {0,1}^S and {0,1}^T, resp., and the map B* -> A* induced by the restriction from T to S is clearly open.

These homeomorphism do not seem to hold true.

This is almost meaningless. :-( I meant: "They are not open."


Now we try to reduce the general situation to the free case above. For this purpose let S be a generating system of A as a Boolean algebra. Denoting the free Boolean algebra over S by Fr(S), the homomorphism Fr(S) -> A (induced by the inclusion S c-> A) is a surjection inducing the map A* -> Fr(S)* which is injective and the topology of A* coincides with the trace topology induced by Fr(S*).

With this in mind, let S=A and T=B systems generating A and B, resp. Then we have the surjection

Fr(A) ->> A

of Boolean algebras yielding the (regular) injection

A* c-> Fr(A)* (*A)

and likewise for B (With regular trace topology is meant).

And unfortunately the claim that A* carries the trace topology fails to be true in general.

Since the surjections B* -> A* are compatible with the mappings (*A) and (*B), the openness of B* -> A* follows from all the facts noted before.

This should complete the proof.


Jose, please let me know whether you could verify all the details and the claim holds true. I appreciate an email in any case.




--
Best wishes,
J.
.

Relevant Pages

  • Re: Continuous maps between boolean/stone spaces
    ... http://planetmath.org/encyclopedia/MHStonesRepresentationTheorem.html) we can assume that the surjection of Stone spaces comes from a homomorphism phi: A -> B of Boolean algebras. ... Let's assume in the first step that A and B are *free* Boolean algebras with basis S and T, resp., with S contained in T. Then the Stone spaces A* and B* are homeomorphic to ^S and ^T, resp., and the map B* -> A* induced by the restriction from T to S is clearly open. ... Denoting the free Boolean algebra over S by Fr, the homomorphism Fr-> A is a surjection inducing the map A* -> Fr* which is injective and the topology of A* coincides with the trace topology induced by Fr. ...
    (sci.math)
  • Re: Continuous maps between boolean/stone spaces
    ... On 08.09.2007 00:10, Jose Capco wrote: ... By Stone's representation theorem we can assume that the surjection of Stone spaces comes from a homomorphism phi: A -> B of Boolean algebras. ... Denoting the free Boolean algebra over S by Fr, the homomorphism Fr-> A is a surjection inducing the map A* -> Fr* which is injective and the topology of A* coincides with the trace topology induced by Fr. ...
    (sci.math)
  • Re: Continuous maps between boolean/stone spaces
    ... "The category of stonean spaces and ... surjection are the same thing (because in the sketch you have a ... I think the assertion that an epimorphism is surjective is equivalent to something like the Hahn-Banach extension theorem in the category of Boolean algebras. ... I think this should help Jose to find out whether his conjecture is true or not, since we are always dealing with duals of Boolean algebras which should be complete I suspect. ...
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  • Re: Cantor Confusion
    ... I give a surjection on all existing paths. ... Each part of an edge may map to a single path, ... The crucial thing in a surjection (and indeed for a mapping) from A to B ...
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  • Re: CANTORs theorem
    ... >> Since that estabishes that there cannot be any surjection from N to ... This is *not* the set M as defined in Cantor's proof for this mapping. ... in the image of the map. ... And that shows that the cardinality ...
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