A while ago I asked a question about recoqnizing CW complexes and got an extremely nice and concrete answer. However, I am still interested in a more general treatment of this and therefore pose the following closely related question:

Assume that $X$ is an $n−1$ dimensional finite CW complex. and assume that $X'$ is given as a a set by the disjoint union of $X$ and a single open cell $e$ of dimension $n$. I.e. $e$ is an open subspace of $X'$ homeomorphic to the open $n$ disc (and of course $X$ is homeomorphic to the complement). Also assume that $X'$ is compact Hausdorff.

Question: Are there some natural topological conditions to further put on $X'$ such that it follows that $X'$ is in fact a CW-complex with $X$ as a sub-CW complex and $e$ a single new cell?

Remark: The fact that $X$ is such is equivalent to whether or not one can modify the homeomorphism of $e$ to the open unit disc such that the inverse extends to the closed unit disc.

Ideas on answers (but I have no proofs or counter examples in any of the cases):

1) $X'$ is locally contractible.

2) $X'$ has the homotopy type of a CW complex.

3) The Combination of 1) and 2).

4) $X'$ is homeomorphic to a CW complex

The last is weird, but it is not clear to me that even this is enough!

Any ideas, counter examples, or references?

  • $\begingroup$ Uh, if X' is homeomorphic to a CW complex, it is a CW complex. $\endgroup$ – Harry Gindi Mar 25 '10 at 17:26
  • $\begingroup$ It is a CW complex, but does it follow that it meets his additional criteria? It has to be a CW complex using the partition into cells he described. $\endgroup$ – Michael Benfield Mar 25 '10 at 18:04
  • $\begingroup$ But anyway, unless I'm misunderstanding something, the boundary of e has to be attached to stuff in X, and that does not happen in your construction, so it doesn't make a CW complex considering e as an open cell. Right? $\endgroup$ – Michael Benfield Mar 25 '10 at 18:15
  • $\begingroup$ I have now rewritten the question to contain the comment about compact Hausdorff in the assumptions before the question. This is more natural to me. $\endgroup$ – Thomas Kragh Mar 25 '10 at 18:32

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