TITLE: Can the product of two $\mathsf Y$'s be embedded in 3-space?
QUESTION [12 upvotes]: Let $Y$ denote the space homeomorphic to the (sans serif) letter $${\huge\mathsf Y}$$ or, equivalently, the space of three closed intervals glued together at one endpoint. Consider the space $Y\times Y$. Here is my attempt at a drawing:

What I drew is not embedded in $\mathbb R^3$ -- it intersects itself. 

Is there any topological embedding $Y\times Y\to\mathbb R^3$? 
If not, why not? 
What is the strategy for attacking problems like this?

REPLY [9 votes]: The product space consists of 9 rectangular sheets, glued together along two of the edges of each sheet. There are 6 glue-edges where three sheets meet, and all of these edges meet at a center point.
Assume we have an embedding and intersect it with a sphere around the center point that is small enough not to contain any points on the boundary of the embedding.
There are 6 points on the sphere that represents glue-edges (just pick an arbitrary one if the glue-edge crosses the sphere multiple times), and the sheets themselves cross the sphere in curves between these points.
Together, the whole figure drawn on the sphere is the complete bipartite graph $K_{3,3}$ -- but that is well known not to be planar, so it can't be drawn on a sphere without intersections!