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Re: Stone Cech
Posted:
Mar 23, 2013 2:52 AM
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On Mar 22, 10:27 pm, Butch Malahide <fred.gal...@gmail.com> wrote:
> On Mar 22, 10:52 am, David C. Ullrich <ullr...@math.okstate.edu>
> wrote:
>
>
>
>
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> > I just realized I have no idea whatever how to write
> > a correct algorithm here, even without worrying
> > about efficiency.
>
> > An equivalence relation on the 2n endpoints
> > determines a compactification. It also determines
> > a graph, where the vertices are the equivalence
> > classes of endpoints and the edges are the
> > original segments. Earlier, thinking that I was
> > thinking about deciding when two of these
> > compactifications are homeomorphic, I was
> > actually thinking about how to determine
> > whether two of the graphs are isomorphic.
> > That's a strictly weaker condition... I have
> > no idea how I'd check whether two equivalences
> > gave homeomorphic compactifications.
>
> You mean, non-isomorphic is a weaker condition than non-homeomorphic.
>
> The remedy for that is to stick to graphs with no vertices of degree
> 2. (OK, you have to make an exception for the graph consisting of a
> single vertex of degree 2, corresponding to the one-point
> compactification of R.)
Oops, I was thinking of connected graphs. In general, you want to work
with graphs with the property that any vertex of degree 2 is the only
vertex in its component. OK, so you have graphs G and H that you want
to test for homeomorphism. Remove vertices of degree 2 (and unify the
two edges that were incident with that vertex) until you get graphs G'
and H', homeomorphic to G and H respectively (G and H are subdivisions
of G' and H') and having the aforesaid property; and then test G' and
H' for isomorphism. So homeomorphism testing (of graphs) is no harder
than isomorphism testing.
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