I'm having a bit of difficulty understanding the following proof from Lee's An Introduction to Topological Manifolds:

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I don't understand the deduction that since $e_0\cap e \subseteq \overline{e}\setminus e$ and $\overline{e}\setminus e$ is contained in the union of finitely many cells of dimension less than $n$, that $e_0\cap e$ is empty since $e_0$ is an $n$-cell.

Why is this exactly?


By defintion (the definition I learned) two cells of a CW complex are either disjoint or equal. Hence if you have cells of different dimensions they are disjoint. Since the author proved that $e_0\cap\bar{e}$ is contained in the finite union of cells of dimension less than the one of $e_0$, $e_0$ has empty intersection with each of them, hence with the union, hence the conclusion.


Thomas has already explained the answer to your question. For my answer, it does not answer your question directly but it provides another reason why each $n$ -cell $e_\alpha^n$ is open in $X$.

You already know that your $n$ -cells are attached via a characteristic map $\varphi_\alpha : S^{n-1} \to Y$. $Y$ now is the space that when you attach $n$ -cells $e_\alpha^n$ you get $X$. Now each $e_\alpha^n$ has its characteristic map $\Phi_\alpha$ which is defined to be the composition

$$D_\alpha^n \hookrightarrow Y \sqcup D_\alpha^n \to Y \sqcup D_\alpha^n/\sim $$

where the quotient space is just $X$. Recall that $\sim$ is the smallest equivalence relation generated by the subset of all pairs $(x,\varphi(x))$ for $x \in S^{n-1}$ in

$$(Y \sqcup D_\alpha^n) \times (Y \sqcup D_\alpha^n).$$

Now it is not hard to see that a subset $A$ of $X$ is open iff $\Phi_\alpha^{-1}(A)$ is open in $D_\alpha^n$ for each characteristic map $\Phi_\alpha$. Now suppose you have an $n$ -cell $e_\alpha^n$. Then if $\alpha \neq \beta$, then it is clear that $\Phi^{-1}_\beta(e_\alpha^n ) = \emptyset$ that is open in $D_\beta^n$. Now if $\beta = \alpha$, this is trivial to see because we would have that $\Phi$ is a homeomorphism of the interior of $D_\alpha^n$ and $e_\alpha^n$.

Is it clear to you why an $n$ - cell is open now?

  • $\begingroup$ Yes, thank you. $\endgroup$ – Holdsworth88 Aug 17 '12 at 14:17

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