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One of axioms of the definition of a CW complex says that the boundary of a cell in contained in a finite union of cells of low dimension. Here, all cells are open cells.

My question is: Given a CW complex, is the boundary of its cells is exactly a finite union of cells of low dimension?

Thanks a lot!

2 Answers2

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No, that is not part of the definition.

And it is easy to construct an example which satisfies the definition and so is a CW complex, but which fails to satisfy that condition: let the $1$-skeleton be homeomorphic to the circle with one $0$-cell $x$ and one $1$-cell $\gamma$; pick a point $y \ne x \in \gamma$; then attach a $2$-cell using the constant attaching map which takes the entire boundary circle to the point $y$.

Lee Mosher
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My understanding is that this does not have to be the case. In particular the attaching map $\psi\colon D^n\to X^{n-1}$ does not have to be a homeomorphism when restricted to $\partial D^n = S^{n-1}$, so strict containment is possible.

Rylee Lyman
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