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