On 3/18/2013 10:06 AM, David C. Ullrich wrote: > On Sun, 17 Mar 2013 19:11:00 -0500, fom <fomJUNK@nyms.net> wrote: > >> On 3/17/2013 8:35 AM, David C. Ullrich wrote: >>> On Sun, 17 Mar 2013 01:28:41 -0700, William Elliot <email@example.com> >>> wrote: >>> >>>> On Fri, 15 Mar 2013, David C. Ullrich wrote: >>>>> On Thu, 14 Mar 2013 19:49:08 -0700, William Elliot <firstname.lastname@example.org> >>>>> wrote: >>>>> >>>>>> Let (g,Y) be a Cech Stone compactification of S. >>>>>> If (f,X) is a compactification of S, does X embed in Y? >>>>> >>>>> Isn't this clear from the universal property of the S-C >>>>> compactification? >>>> >>>> No. All I get is that there's a closed continuous surjection h >>> >from Y onto X for which h|g(S) is injective. >>> >>> I didn't read the question carefully, sorry. Sort of assumed >>> it was what would seem like a sensible question regarding >>> the universal property of the S-C compactification. >>> >>> The answer to the actual question is no, X does not embed in Y, >>> or at least "surely not - the defining property of the S-C >>> compactification simply has nothing to do with spaces >>> embedding in Y". >>> >>> What's true is that X is a _quotient space_ of Y. >>> >>>> >>>>>> If (g,Y) is a compactification of S and >>>>>> for all compactifications (f,X), X embeds in Y >>>>>> is (g,Y) a Stone Cech compactification of S? >>> >>> Similarly here - I misread the question as something about >>> whether another compactification sharing the same >>> universal property as the S-C compactification must >>> be homeomorphic to the S-C compactification (the >>> answer to _that_ is yes, and the proof starts as >>> I suggested). >>> >>> There's simply no reason to think that the answer to >>> the question you ask is yes, unless possibly it's yes >>> vacuously - I can't imagine an example of (g,Y) that >>> has the property you assume here. >>> >>> In particular, when you ask "If (g,Y) has property P, >>> must (g,Y) be a S-C compactification of S?" I sort of >>> assumed that property P must be a property that >>> the S-C compactification actually _satisfies_. >>> That's simply not so here - it's not true that every >>> compactification of S embeds in the S-C compactification. >>> >> >> So, how does your last statement >> reconcile with the text I quoted >>from Munkres that describes Stone-Cech >> compactification as maximal "in some >> sense". > > (Under suitable hypotheses; being an analyst > assuming locally compact Hausdorff is fine > with me:) > > The S-C of X, let's call it bX, is maximal > in the sense that every compactification > is a _quotient space_ of bX. That's > simply a totally different thing from > saying every compactification embeds > in bX, which is not true (or if it is true, > it's true just by accident). >
Right. Within Munkres' description it is expressly mentioned that the construction used *satisfies* the imbedding theorem. So, there is no reason to think that imbeddings are generally in correspondence with the quotient maps involved without an actual proof as such.
The notion of imbedding is logically prior to that (particular) construction.
> Consider R, the real line. The space bX > consists of R with a huge amount of > fuzzy sttuff tacked on at the ends. > The one-point compactification of R > is obtained from bX by taking all the points > other than points of R and "identifying" > them to a single point. > > Otoh does the one-point compactification > of R embed in bR? I don't know for sure, > but I doubt it. And it _is_ clear that it doesn't > embed in the relevant sense: > > The question is whether bR contains > something homeomorphic to S^1. I doubt > it.
Right. There is some R^n into which the one-point compactification imbeds topologically. At most, n=2k+1=3 if I am not mixing up unrelated theorems.
But there is no reason to think that bR has the needed structure to which you refer.
> But it's clear that there is no S^1 in bR > that consists of R plus one more point, > which is what I mean by saying there's no > "relevant" embedding of S^1 in bR. If > there is an S^1 in bR it's just a random > circle sitting somewhere in that fuzzy > stuff, that really has nothing to do with > the fact that bR is bR. > > Hmm. This gives a better answer to the OP's > question about why a compactiification is defined > as an ordered pair including an embedding, instead > of just as a topological space. If a compactification > of X is a topological space then the notion of one > compactification embedding in another includes > embeddings that are simply irrelevant to X. > > Otoh it's easy to give a definition that captures > what it "really means", or should mean, for one > compactification to embed in another. Say > (g,A) is a compactification of X (so g : X -> A > is a map such that etc.) Say (h, B) is another > compactification of X. Then (g,A) "compactification- > embeds" in (h, B) if there exists an > embedding f : A -> B such that f(g(x)) = h(x) > for all x in X. (If we pretend that X s literally > a suubset of A and of B then the condition is that > f fix every point of X.) > > Which come to think of it surely never happens > unless f is a homeomophism: f(A) is compact, > hence closed, and contains X, so f(A) is dense, > hence f(A) = B and standard blah blah shows > the inverse of f is continuous. > > So. Given the _relevant_ notion of one compactification > embedding in another, it never happens unless the > two compactifications are equivalent. Embedding > is simply not what we're talking about when we talk > about bX. >
This last paragraph sounds like a nod toward some items in Willard concerning uniform spaces.
Now that I look, William might wish to check section 41. It addresses imbeddings of Tychonoff spaces into compact spaces and how each such imbedding corresponds with a compatible proximity.
Compactifications corresponding with proximities are called Samuel compactifications.
The section concludes with an analysis of when a Samuel compactification is the Stone-Cech compactification.
So, at least this is where one is looking at imbeddings and comparability of different compactifications.
Thank you for your remarks. Once you pointed out the quotient space relation, the answer to my question became obvious.