The Question
Consider the set of all non-orientable \(n\)-dimensional pseudomanifolds. Let \(X = \cup_i X^i\), \(X^0 \subset X^1 \subset X^2 \subset \cdot\cdot\cdot\) be an element from this set, with \(X^i\) describing the \(i\)-th dimensional skeleton of the CW-complex structure. Similarly, let \(Y = \cup_i Y^i\), \(Y^0 \subset Y^1 \subset Y^2 \subset \cdot \cdot\cdot\) be another element of this set. Let \(|X^i|\) denote the number of \(i\)-cells that were attached to \(X^{i-1}\) to form \(|X^i|\), and define \(|Y^i|\) similarly. Denote by \(k\) the largest integer such that \(|X^k| \neq |Y^k|\). Let \(X < Y\) if \(|X^k| < |Y^k|\), this defines a poset structure on our set of non-orientable \(n\)-dimensional pseudomanifolds.

Describe the minimal elements of this poset.
Investigate how many dimensions of Euclidean space are needed for the minimal elements to be embedded in Euclidean space. Is this constant for the family of minimal pseudomanifolds? If not, we can further distinguish between our minimal nonorientable pseudomanifolds by saying that the one which requires the lowest dimension to be embedded in is the smallest.

Real Projective Plane
The Real Projective Plane (\(RP^2\)) in its canonical CW-complex form (1 0-cell, 1 1-cell, and 1 2-cell that is wrapped around the 1-cell twice I think) is not actually a pseudomanifold, because the 2-cell's boundary is not homeomorphically mapped onto the 1-cell. (Also more trivially, pseudomanifold requires each 1-cell to be the boundary of exactly 2 2-cells) Can we give the real projective plane an appropriate pseudomanifold CW-structure?

Question
Are we asking the right question here? Perhaps we shouldn't be defining the partial order based on the CW-complex structure given, but rather something more 'innate' that defines a manifolds complexity.... it seems like even the 'minimal number of Euclidean dimensions needed to embed object' is a better measure for complexity and makes more sense

Find an appropriate CW-structure for the Klein Bottle
That makes it a pseudomanifold. If I think about it; obviously if I triangulate the surface with really small triangles it will work, but that's not the smartest way to go about it probably. There's definitely an easy CW-structure I can give it.

Solution 1
Just start with the square (with opposite edges identified, one parallel one antiparallel) and chuck a smaller square inside the square, extend all the edges so they wrap around. This divids the square into \(4\) 2-cells, and it all works out. Beautiful pseudomanifold.

Jordan-Brouwer
Somehow; it works like this in my head. Any 2-manifold; if embedded into 3-dimensions, has to split it into two regions by some application of Jordan Brouwer. Then that means that it encloses a '2 dimensional' hole, and so its top homology group is infinite cyclic, and it must be oriented. Extend this proof?

Somehow you'd have to prove that every embedded manifold into 3 dimensions is homeomorphic to a sphere? Is this even true?