categories: Open maps of locales

["Dr. P.T. Johnstone" <P.T.Johnstone@dpmms.cam.ac.uk>]

A couple of weeks ago, Walter Tholen asked whether, if a map of
locales f: X --> Y has the property that pulling back along f
preserves closures of sublocales, then f is necessarily open.
(The converse is true, and the corresponding condition for spaces
is equivalent to openness.) I found a counterexample, and posted
it on this mailing list. Subsequently Walter asked me a
supplementary question: if f stably has the property above
(i.e. all pullbacks of f have the property), is it necessarily
open? I now have a positive answer to this question, which may
be of interest to people who read my earlier posting; so here it is.

To prove the result for spaces, one argues as follows: given an open
U \subseteq X, consider the complement V of the image f_!U in Y. The
inclusion V \subseteq cl V is dense, so it must pull back to a dense
inclusion; but f^*V \cap U is empty, so this implies f^*(cl V) \cap U
is empty, i.e. cl V is disjoint from f_!U. So V = cl V is closed, and
hence f_!U is open.

The reason why this argument doesn't work for locales is, of course,
that f_!U need not have a complement in the lattice of sublocales of Y.
However, if we know that f_!U is a closed sublocale of Y, then the
argument works exactly as for spaces, since closed sublocales are
complemented. So the trick is to pull back along a morphism
\tilde{Y} --> Y such that the pullback of f_!U becomes closed;
specifically, we take the frame {\cal O}(\tilde{Y}) to be the subframe
of the assembly of Y (the frame of all nuclei on Y) generated by the
closed nuclei together with the nucleus j corresponding to f_!U.

We need an explicit description of the elements of this frame: but it
is easy to see that they are all nuclei of the form
(j \cap c(V)) \cup c(W), where V and W are opens of Y (and we may as
well assume V \supseteq W, i.e. c(V) \geq c(W)). Then we observe that
j acquires a complement in this frame iff there exist V and W such that

0 = j \cap ((j\cap c(V))\cup c(W)) = j\cap c(V)  and
1 = j \cup ((j\cap c(V))\cup c(W)) = j\cup c(W)

(where 0 and 1 denote the bottom and top elements of the frame of
nuclei). But this is equivalent to saying that j is already open as
a nucleus on {\cal O}(Y). In other words, if f_!U becomes open when
pulled back to a sublocale of \tilde{Y}, then it was already open as
a sublocale of Y.

Of course, it needs checking that, in the square

  \tilde{U} -------> \tilde{f_!U}
     |                  |
     |                  |
     |                  |
     v                  v
  \tilde{X} -------> \tilde{Y}

obtained by pulling back along \tilde{Y} --> Y, the top edge is an
epimorphism, so that \tilde{f_!U} is the image of \tilde{U} under
\tilde{f}. But this follows easily from the fact that, for any
g: Z --> Y, \tilde{Z} is the locale corresponding to the frame
obtained from {\cal O}(Z) by `declaring g^*(f_!U) to be closed'.
For we have f^*(f_!U) \supseteq U, and hence \tilde{U} --> U is
an isomorphism, as is \tilde{f_!U} --> f_!U.

Peter Johnstone