stable flatness and exponentiation

stable flatness and exponentiation

[Martin Escardo]

I have three questions:

Question 1: Is there an intrinsic characterization of the stably flat
maps of locales, i.e., those continuous maps f: X->Y such that

   id_Z x f: Z x X -> Z x Y

is flat for all Z?

(Recall that f: X->Y is flat iff the right adjoint f_*: OX -> OY of
f^*: OY -> OX preserves finite joins. Preservation of merely the empty
join amounts to density, and hence flatness is a rather strong form of
density. For example, the embedding of a completely regular locale
into its Stone-Cech compactification is strongly dense in this sense.)

I don't think all flat maps are stably flat, but I may be
mistaken. Specifically, for a locale X let JX denote the spectral
locale defined by

   OJX = ideals of OX.

Then there is a sublocale embedding eta: X->JX defined by

   eta^*(I) = join I,

which is known to be flat.

Question 2: Is eta: X->JX stably flat for every X?

If so, then compact Hausdorff locales would be exponentiating. Here a
locale Y is called exponentiating iff the exponential Y^X exists for
every X. Recall also that X is exponentiable iff the exponential Y^X
exists for every Y, and that this is the case iff the frame OX is a
continuous lattice.

Now, JX is exponentiable because OJX is an algebraic lattice. We claim
that if eta_X were stably flat, then for every compact Hausdorff
locale Y, the exponential Y^JX would have the universal property of an
exponential Y^X with respect to a suitably defined evaluation map and
a suitable construction of exponential transposition.

Define e: Y^JX x X -> Y as the restriction of the original evaluation
map ev: Y^JX x JX -> Y, that is

             id x eta                ev
   Y^JX x X ----------> Y^JX -> JX -------> Y.
            ------------------------------>
                          e

Now, to show that the pair (Y^JX,e) has the universal property of an
exponential Y^X, given f: Z x X -> Y, we have to construct a transpose
f': Z -> Y^JX such that e(f' x id_X) = f.

Consider the diagram


              id_Z x eta_X
       Z x X --------------> Z x JX
          \                   .
           \                 .
            \               .
             \             .
              \           .
            f  \         .  f''
                \       .
                 \     .
                  \   .
                   v v
                    Y.

If eta_X were stably flat, then, by "Joyal's Lemma", which says that
compact Hausdorff locales are orthogonal to flat embeddings, there
would be a unique f'' making the diagram commute.

Then its Y^JX-transpose f': Z -> Y^JX with respect to the original
evaluation map would give the unique required Y^X-transpose of our
given f: Z x X -> Y with respect to our constructed evaluation map, as
an easy calculation shows, using the universal property of Y^JX with
respect to the original evaluation map.

This shows that if Question 2 had a positive answer then compact
Hausdorff locales would be exponentiating.

Question 3. But surely compact Hausdorff locales cannot possibly be
exponentiating, can they?

(These questions make sense for topological spaces too. )

MHE

Re: stable flatness and exponentiation

[Martin Escardo]

Martin Escardo writes:
 > Question 3. But surely compact Hausdorff locales cannot possibly be
 > exponentiating, can they?
 >
 > (These questions make sense for topological spaces too. )

Clarification:

Alex Simpson points out, in a reply not sent to the list, that in the
ambient category of topological spaces, compact Hausdorff spaces are
not exponentiating. For example the exponential 2^(N^N) doesn't exist,
where 2 is the two-point discrete space and N is the discrete space of
natural numbers, and, of course, the exponential N^N does exist. In
fact, Alex emphasized this years ago, when I subjected him to the
ideas presented in the previous message.

The point is that the locale product of two (sober) topological spaces
doesn't coincide with the topological product, and hence exponentials
are potentially different, as they are defined with respect to
products.

You may say: well, in any case, it is a fact that a sober space is
exponentiable in Top iff it is exponentiable in Loc, so Alex's
counter-example should work in Loc too.  But then I reply: this
coincidence has to do with the fact that the locale product coincides
with the topological product if one of the factors is locally
compact. In our case, because the exponent is NOT necessarily locally
compact, this coincidence fails. We are considering exponentiating
rather than exponentiable spaces. So, in principle, it may be that
compact Hausdorff locales are exponentiating in Loc - although I very
much doubt that this would be the case, as should be clear from the
previous message. The point is that I just don't know, and I am
looking forward to be enlightened after my failed attempts to decide
the question either way.

In light of Alex's observation, my conclusion to the previous message
should have been:

> Question 3. But surely compact Hausdorff locales cannot possibly be
> exponentiating, can they?

> It would be rather amazing if they were, because compact Hausdorff
> spaces are not exponentiating in the category of topological
> spaces. But the known arguments in the case of topological spaces
> don't seem to apply here.

One further comment: it is plausible that eta: X->JX is stably flat
for X exponentiable. If Alex's topological counter-example applies to
locales, then eta_X is not stably flat for X=N^N.

MHE