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$\newcommand{\R}{\mathbf R}$ For a cell $e$ in a CW complex, we write $\partial e$ to denote $\bar e-e$. Note that $\partial e$ may not be the topological boundary of $e$ in $X$.

A CW complex structure on a topological space $X$ is said to be regular if each cell in $X$ admits a characteristic map which is a homeomorphism.

I am trying to show that

Conjecture. Let $e^k_\alpha$ be a $k$-cell and $e^n_\beta$ be an $n$-cell in a regular CW complex such that $k<n$. If $e^k_\alpha\cap \partial e^n_\beta\neq \emptyset$, then $e^k_\alpha\subseteq \partial e^n_\beta$.

My reason of thinking that the above conjecture is true is Theorem 2 which is proved below.

Theorem 1. There is no embedding of $S^n$ in $\R^n$.
Proof. By the Borsuk-Ulam theorem, any continuous map $S^n\to \R^n$ admits a pair of antipodal points in $S^n$ which have the same image. Thus there does not exist an injective continuous map $S^n\to \R^n$, proving the theorem.

Invariance of Domain.
Let $f:U\to \R^n$ be an embedding of an open subset $U$ of $\R^n$ into $\R^n$. Then $f(U)$ is open in $\R^n$.

Theorem 2. Let $e_\alpha^{n-1}$ and $e_\beta^n$ be $(n-1)$ and $n$-cells respectively in a regular CW complex. Assume that $e^{n-1}_\alpha\cap \partial e^{n}_\beta\neq \emptyset$. Then $e^{n-1}_\alpha\subseteq \partial e^n_\beta$.
Proof. Since $\partial e^n_\beta\subseteq X^{n-1}$ and $e^{n-1}_\alpha$ is open in $X^{n-1}$, we have $e^{n-1}_\alpha\cap\partial e^n_\beta$ is open in $\partial e^{n}_\beta$. Note that $e^{n-1}_\alpha\cap \partial e^{n}_\beta\neq \partial e^n_\alpha$. For otherwise we would have an embedding of $S^{n-1}\cong \partial e^n_\beta$ into $\R^{n-1}\cong e^{n-1}_\alpha$, contrary to Theorem 1. This helps us invoke Invariance Of Domain to conclude that $e^{n-1}_\alpha\cap \partial e^n_\beta$ is open in $e^{n-1}_\alpha$. But since $\partial e^n_\beta$ is compact, we also know that $\partial e^n_\beta\cap e^{n-1}_\alpha$ is closed in $e^{n-1}_\alpha$. Therefore if $e^{n-1}_\alpha\cap \partial e^n_\beta\neq \emptyset$, the connectedness of $e^{n-1}_\alpha$ forces $e^{n-1}_\alpha\subseteq \partial e^n_\beta$.

Here is my attempt in trying to prove the conjecture. Let $e^n_\beta$ be an $n$-cell in a regular CW complex and let $e^{n-1}_1, \ldots, e^{n-1}_p$ be all the $(n-1)$-cells which intersect $\partial e^n_\beta$. Thus by Theorem 2 we have $e^{n-1}_i\subseteq \partial e^n_\beta$ for all $i$. Further, since $\partial e^n_\beta$ is compact, we can in fact write $\bar e^{n-1}_i\subseteq \partial e^n_\beta$. Therefore $C:=\bigcup_{i=1}^p \bar e^{n-1}_i$. Now $C$ is a closed subset of $\partial e^n_\beta$. If I can show that $C$ is the whole of $\partial e^n_\beta$ then I can use induction of the dimension of a cell to prove my conjecture. So on the contrary assume that $C$ is properly contained in $\partial e^n_\beta$. Then $O=\partial e^n_\beta\setminus C$ is a non-empty open subset of $\partial e^n_\beta$. Thus $O$ is homeomorphic to an open subset of $S^{n-1}$. Now $O$ is covered by "parts of" finitely many cells of dimension less than $n-1$. This seems intuitively impossible. This also raises the following question:

Question. Can we give $\R^n$ an $(n-1)$-dimensional (possibly infinite) CW structure?

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Concerning your question, the answer is no. Suppose that $\mathbb{R}^n$ has an $(n-1)$-dimensional CW structure (which by the way must be infinite, since otherwise it would imply that $\mathbb{R}^n$ is compact). Then, let $e$ be a top dimensional cell (with dimension $n-1$). Take $p \in e - \partial e$. Then, $p$ has an open neighbourhood homeomorphic to some open subset of $\mathbb{R}^{n-1}$. However, since our space is $\mathbb{R}^n$, this open set must be also an open subset of $\mathbb{R}^n$. But an standard argument using invariance of domain shows that no open set of $\mathbb{R}^k$ is homeomorphic to an open set of $\mathbb{R}^l$ if $k \neq l$.

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  • $\begingroup$ Oh so we are merely using the fact that a top dimensional cell in open. Nice. $\endgroup$ – caffeinemachine Feb 9 '16 at 16:21
  • $\begingroup$ Yes. By the way, your proof of your "conjecture" sounds OK to me. $\endgroup$ – TopologicalGuy Feb 9 '16 at 16:23
  • $\begingroup$ I am not able to formally put the last line where I say an open set in $S^{n-1}$ is a union of "parts of" finitely many ... Can you please help me with that? $\endgroup$ – caffeinemachine Feb 9 '16 at 16:54
  • $\begingroup$ Just observe that $\partial e^n_\beta$ is itself a CW complex. Now you can argue as before: a point in $O$ must have an open neighbourhood homeomorphic to an open neighbourhood of an euclidean space of dimension less than $n-1$, which at the same time is an open neighbourhood of $S^{n-1}$... $\endgroup$ – TopologicalGuy Feb 9 '16 at 17:31
  • $\begingroup$ I am unable to see that $\partial e^n_\beta$ is a CW complex. By $\partial e^n_\beta$ I mean $\bar e^n_\beta-e^n_\beta$. Can you please elaborate as to how can we view this is as a cell complex. $\endgroup$ – caffeinemachine Feb 9 '16 at 17:37

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