Skip to main content
19 events
when toggle format what by license comment
May 4, 2016 at 4:41 vote accept Włodzimierz Holsztyński
Apr 27, 2016 at 15:39 answer added Lasse Rempe timeline score: 1
Apr 30, 2015 at 7:46 history edited Włodzimierz Holsztyński CC BY-SA 3.0
Clearer? (Also, 1 typo fixed).
Apr 13, 2015 at 17:38 comment added Włodzimierz Holsztyński @MathieuBaillif -- very well said! Thank you Mathieu.
Apr 13, 2015 at 16:40 comment added Mathieu Baillif I think the OP wants to know whether being a universal image for Hausdorff compact spaces implies (and thus is equivalent to) being snake-like (in the class of compact connected spaces).
Apr 13, 2015 at 13:36 comment added Gabriel C. Drummond-Cole @WłodzimierzHolsztyński I still think I'm misunderstanding something. Why isn't $X=S^1=Y$ a counterexample? Or you want to find a space that accepts a universal map? Or...?
Apr 13, 2015 at 11:01 comment added Włodzimierz Holsztyński @DavidWhite -- I misunderstood of what you said. I thought that you were objecting to the gray field, not to what was under the gray field, i.e. that you didn't like the subtitle "D E F I N I T I O N(s)" itself.
Apr 13, 2015 at 10:55 history edited Włodzimierz Holsztyński CC BY-SA 3.0
TeX typo
Apr 13, 2015 at 9:17 comment added Włodzimierz Holsztyński @GabrielC.Drummond-Cole -- your edit is fine, thank you. Your format is more pleasing to an eye, especially due to the existing standards. My format allows to see the sense faster (granted that you do not feel shocked by its being optically irregular, that you let it be). It's like x++ to a Pascal programmer who is used to x:=x+1.
Apr 13, 2015 at 9:07 comment added Włodzimierz Holsztyński @DominicvanderZypen -- when $\ Y\ $ is snake-wise (etc) then the respective $\ p\ $ exists. Thus it is vital for the question to ask about spaces $\ Y\ $ which are not snake-like continua.
Apr 13, 2015 at 8:37 comment added Dominic van der Zypen Actually we don't have to know what snake-like is: if I understand correctly we just need to answer this question: Suppose $X,Y$ are compact Hausdorff and connected, and $f,v:X\to Y$ are continous with $v$ surjective. Then: is there always $p\in X$ sucht that $f(p) = v(p)$?
Apr 13, 2015 at 8:25 comment added Gabriel C. Drummond-Cole @WłodzimierzHolsztyński I hope you don't find my edits intrusive; I made them because I was genuinely confused about the $u\ f$ and $v\ f$ notation. I apologize if they constitute an overreach, I know you are sometimes particular about wording.
S Apr 13, 2015 at 6:39 history suggested Gabriel C. Drummond-Cole CC BY-SA 3.0
fixed a couple of typos and clarified that the concatenations uf and vf each referred to two separate functions
Apr 13, 2015 at 6:29 review Suggested edits
S Apr 13, 2015 at 6:39
Apr 13, 2015 at 4:10 comment added Todd Trimble @DavidWhite It's there by intention, as several recent discussions have borne out. Probably no one but the author understands why he type-sets that way, but I think it would be wise not to pursue this here and now. The mathematics looks interesting.
Apr 13, 2015 at 3:00 comment added Włodzimierz Holsztyński Confused?? There are no special rules about the contents of the gray fields that I know of.
Apr 13, 2015 at 2:58 history edited Włodzimierz Holsztyński CC BY-SA 3.0
clearer?
Apr 12, 2015 at 19:43 comment added David White I am a bit confused by the type-setting. Is that what is supposed to be in the grey box?
Apr 11, 2015 at 17:29 history asked Włodzimierz Holsztyński CC BY-SA 3.0