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I'm reading "Memento on cell complexes" and the following is supposed to be a counter example of a cell complex because $e^1$ is not homeomorphic to the open segment:

enter image description here

To me this looks like a deformed disk and is therefore homeomorphic to $D^2$. Why is $e^1$ not homeomorphic to the open segment? Or is it missing a second black dot to denote touching of $e^1$ at 12 o'clock?

Also, I don't see why the following isn't a cell complex either. The boundary of $e^2$ maps into $X^1$. The inductive construction of a cell complex requires exactly that.

enter image description here

Thanks for your help!

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It looks like in $C_4$ the cell $e^1$ is intended to touch itself and enclose the tear drop. In $C_5$ the labeling indicates that there is only one $0$-cell and two $1$-cells and $e_{2}^1$ is attached in the middle of $e_{1}^1$ which is not allowed. –  t.b. Sep 2 '11 at 11:14
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Another way to say this is ANY CW-complex with a single 0-cell and a single 1-cell must have 1-skeleton homeomorphic to a circle. But the space pictured has a "1-skeleton" homotopy-equivalent to a wedge of two circles. –  Ryan Budney Sep 2 '11 at 16:32
    
OK, so in case one, there should indeed be another black dot. And in case two, the entirety of $\partial e^2$ should be mapped, not just one dot on it! Thanks for your help! –  Matt N. Sep 3 '11 at 9:46
    
@RyanBudney and t.b. , is there a reason you left comments only and not answers? It looks like this question is resolved but I can't tell for sure because it hasn't been officially marked as such. –  isomorphismes Oct 5 '12 at 22:36

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