Andy wrote: << More simply, take two equilateral triangles, identify all three corners of one with all three corners of the other. This does not have an affine embedding in R^n for all n. This (along with identifying some of the edges of the triangles) is exactly what you've done with the two center triangles of your strip, in your more complex example. Even simpler, in one dimension less, take two line segments, and identify their endpoints. This does not embed in R^n affinely for any n. . . .
The problem is, in a simplicial complex any two simplices that intersect must do so in one common face (of any dimension). So two 2-simplices that intersect in all three 0-faces doesn't qualify. Ditto, of course, for two 1-simplices that intersect in both their endpoints. --Dan