Mike Reid writes:
<<
[I asked]:
> Determine whether or not the topological space Y x Y
> (the cartesian product of two Y-shaped spaces, each
> the union of 3 closed intervals) can be embedded as
> a subset of 3-space.
it cannot be embedded. Y x Y is homeomorphic to the cone over
the graph C_3 x C_3 . if there were such an embedding, intersecting
with a small sphere centered at the vertex (the point (y_0, y_0)
in Y x Y , where y_0 is the branch point of Y ) would give an
embedding of C_3 x C_3 in the sphere. this is impossible, because
this graph is non-planar.
>>
Except for the fact that I would say the graph that Y x Y is a
cone over is K(3,3), that is exactly my conclusion, and my reasoning.
(By the Leibniz rule,
bd(Y x Y) = bd(Y) x Y u Y x bd(Y),
(with agreement on bd(Y) x bd(Y)) -- which is homeomorphic to K(3,3).)
(This is quite clearly valid for Y x Y embedded either smoothly
or as a simplicial complex. There may be some bothersome
technical details to check, however, if the embedding is only
known to be continuous.)
--Dan
