I just noticed this question. I think I answered essentially the same question from the OP by email, but I'll repeat the answer here for public consumption.
Ordinarily John M. Lee's book "Introduction to Topological Manifolds"
is a model of clarity,
Thank you!!
but on p. 131 I am very confused by the following text
"Given a cell complex (X, E) the open cells are typically just called
the cells of X. Be careful: although each e contained in E is an open
cell meaning that it is homeomorphic to B^n for some n, it is not
necessarily an open subset of X. By the closed map lemma and
proposition 2.30 (p. 30) the image of a characteristic map for e is
equal to e^bar so each cell is precompact in X; but its closure might
note be a closed cell because the characteristic map need not be
injective on the boundary."
What does 'its' refer to? How could e^bar NOT be closed? What does
non-injection on the boundary have to do with anything?
The point of the phrase you quoted, "its closure might note be a closed cell," is that the closure of $e$ might not be a closed cell, not that it might not be closed. Of course it's closed. But to be a closed cell, it would have to be homeomorphic to a closed Euclidean ball, and it might not be if the characteristic map isn't injective on the boundary. A good example to keep in mind is the map $q$ of Example 4.55: It maps the open ball homeomorphically onto an open subset of the sphere (which is therefore an open $n$-cell in the sphere), but it collapses the boundary to a point. So the closure of the open $n$-cell is definitely not a closed $n$-cell.
How could something homeomorphic to the classic open set (B^n) not be
open?
Just take $X=\mathbb R^2$ and $e=\mathbb R \times \{0\}$. Then $e$ is homeomorphic to $\mathbb R$ (and thus to $\mathbb B^1$), so it is an open $1$-cell. But it's not an open subset of $X$.
Jack Lee
