1
$\begingroup$

If $(X, \Gamma)$ is a cell complex and $e \in \Gamma$, then the image of a characteristic map for $e$ equals $\bar{e}$

I'm trying to prove the above which is a statement in Introduction to Topological Manifolds by John Lee

My Attempted Proof: Let $(X, \Gamma)$ be a cell complex and choose an open $n$-cell $e \in \Gamma$ of dimension $n \geq 1$. Let $\phi$ denote the characteristic map for $e$. Since $\phi$ is a characteristic map we have $\phi : D \to X$ for some closed $n$-cell $D \subseteq Y$ (where $Y$ could be any topological space) with the properties that $\phi' : \operatorname{Int}(D) \to e$ which is the restriction of $\phi$ to $\operatorname{Int}(D)$ is a homeomorphism and the restriction of $\phi$ to $\operatorname{Bd}(D)$ denoted by $\phi|_{\operatorname{Bd}(D)}$ maps $\operatorname{Bd}(D)$ into the union of all cells of $\Gamma$ of dimensions strictly less than $n$.

We need to show that $\phi[D] = \bar{e} = \operatorname{Cl}_X(e)$. By continuity of $\phi$ we have $\phi[\operatorname{Cl}_Y(D)] \subseteq \operatorname{Cl}_X(\phi[D])$ and since $\phi$ is a map from a compact space to a Haursdoff space we have that $\phi$ is a closed map and hence $ \operatorname{Cl}_X(\phi[D]) \subseteq \phi[\operatorname{Cl}_Y(D)]$ so we have $ \operatorname{Cl}_X(\phi[D]) =\phi[\operatorname{Cl}_Y(D)]$. Now if $D = \operatorname{Cl}_Y(D)$ (****) and hence $D$ is a closed set in $Y$ we can go ahead and say that $\phi[D] = \operatorname{Cl}_X(\phi[D])$.

Now note that we have $\phi[\operatorname{Int}(D)] =\phi'[\operatorname{Int}(D)] = e$. Observe that by continuity of $\phi$ we have $\phi[\operatorname{Cl}_Y(\operatorname{Int}(D))] \subseteq \operatorname{Cl}_X(\phi[\operatorname{Int}(D)]) = \operatorname{Cl}_X(e) $. Assuming our above assertion holds since $D$ is homeomorphic to $\overline{\mathbb{B}^n} \subseteq \mathbb{R}^n$ with $\operatorname{Int}(D)$ homeomorphic to $\operatorname{Int}(\overline{\mathbb{B}^n}) = \mathbb{B}^n$ and $\operatorname{Cl}_{\mathbb{R}^n}(\mathbb{B}^n) = \overline{\mathbb{B}^n}$ we have $\operatorname{Cl}_Y(\operatorname{Int}(D)) = D$. Now since $\phi$ is a closed map (by the argument in the above paragraph) we have $ \operatorname{Cl}_X(\phi[\operatorname{Int}(D)]) = \operatorname{Cl}_X(e) \subseteq \phi[\operatorname{Cl}_Y(\operatorname{Int}(D))] $, hence we have $\operatorname{Cl}_X(e) =\bar{e} = \phi[\operatorname{Cl}_Y(\operatorname{Int}(D))] = \phi[D]$ as desired.


Now there is a huge flaw in this proof I think at the (****) part as we have no idea that $D = \operatorname{Cl}_Y(D)$ so we can't conclude that $D$ is an closed set in $Y$. Sure $D$ is homoemorphic to a closed subspace of $\mathbb{R}^n$, but homeomorphisms don't preserve closures of improper subspaces.

For example define $f : [0, 1] \to (0, 1)$ by $$ f(x) = \begin{cases} x & \text{if} \ \ x \in (0, 1) \\ 0 & \text{if} \ \ x = 1 \ \ \text{or}\ \ x = 0 \\ \end{cases}$$

Then the restriction of the domain of $f$ to $(0, 1)$ is a homeomorphism, however $\operatorname{Cl}_{[0, 1]} \left((0, 1)\right) = [0, 1]$ but $\operatorname{Cl}_{(0, 1)}((0, 1)) = (0,1)$.

So how can I rectify this into a correct proof? I believe that I may have made this a bit too complicated and the proof is probably a lot simpler.


Side note: The book that I am using Introduction to Topological Manifolds by John Lee does not assume the closed $n$-cell $D$ is embedded in a parent topological space, so there isn't any mention of a topological space $Y$ for which $D \subseteq Y$, however if we don't have such a topological space $Y$, then we'd have $\operatorname{Int}(D) = D = \operatorname{Cl}_D(D)$ because we'd be taking the interior of $D$ in the topological space $D$ as opposed to $Y$, and the largest open set in $D$ is obviously $D$. And we'd arrive at contrdictory stuff like $D$ is homeomorphic to both $\overline{\mathbb{B}^n}$ and $\mathbb{B}^n$.

$\endgroup$
  • $\begingroup$ See the answer here, there's no need to bring $Y$ into the picture: math.stackexchange.com/questions/476582/… $\endgroup$ – leibnewtz Dec 20 '17 at 6:09
  • $\begingroup$ You are indeed making too much out of the concept of a "closed $n$-cell". By definition, a closed $n$-cell is any topological space $D$ that is homeomorphic to $$B^n = \{x \in \mathbb{R}^n \,\bigm|\, |x| \le 1\}$$ So $D$ is compact, and $D$ is the closure of its interior, and various other things. For example, in the second paragraph of your attempted proof you could have written "By continuity of $\phi$ and compactness of $D$, $\phi[D]$ is compact", which would simplify several other later arguments. $\endgroup$ – Lee Mosher Dec 20 '17 at 18:33