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 pick an open cell $e \in \Gamma$. By definition of $\Gamma$ being a cell decomposition of $X$, we have that $e$ has dimension $n \geq 1$.

Now let $\Phi : D \subseteq Y \to X$ be a characteristic map for $e$. (Note that $D$ is a closed $n$-cell and is homeomorphic to $\overline{\mathbb{B}^n}$). Since $D$ is compact (because $\overline{\mathbb{B}^n}$ is) and $\Phi$ is continuous we have $\Phi[D]$ to be compact.

Observe that $\Phi$ is a continuous map from a compact space to a Hausdorff space (by which we mean $\phi[D]$ and not $X$). Hence $\Phi$ is a closed map by the closed map lemma.

By continuity of $\Phi$ we have $\Phi[\operatorname{Cl_Y}(\operatorname{Int_Y}(D))] \subseteq \operatorname{Cl_X}(\Phi[\operatorname{Int_Y}(D)])$ and since $\Phi$ is a closed map we have $\Phi[\operatorname{Cl_Y}(\operatorname{Int_Y}(D))] \supseteq \operatorname{Cl_X}(\Phi[\operatorname{Int_Y}(D)])$, hence $\Phi[\operatorname{Cl_Y}(\operatorname{Int_Y}(D))] = \operatorname{Cl_X}(\Phi[\operatorname{Int_Y}(D)]) = \operatorname{Cl_X}(e)$ since the restriction of $\Phi$ to $\operatorname{Int_Y}(D)$ is homeomorphic to $e$, and we have $\Phi[\operatorname{Int_Y}(D)] = e$.

Now if we assume the following lemma is true:

If $A \subseteq M$ is any subspace that is homeomorphic to a closed set $B \subseteq N$ with $\operatorname{Int_M}(A)$ homeomorphic to $\operatorname{Int_N}(B)$ and $\operatorname{Cl_N}(\operatorname{Int_N}(B)) = B$ then $\operatorname{Cl_M}(\operatorname{Int_M}(A)) = A$

Then if the above holds we then have $D \subseteq Y$ which is homeomorphic to the closed set $\overline{\mathbb{B}^n} \subseteq \mathbb{R}^n$ with $\operatorname{Int_{\mathbb{R}^n}}(\overline{\mathbb{B}^n}) = \mathbb{B}^n$ which is homeomorphic to $\operatorname{Int_Y}(D)$ and we have $\operatorname{Cl_{\mathbb{R}^n}}(\operatorname{Int_{\mathbb{R}^n}}(\overline{\mathbb{B}^n})) = \operatorname{Cl_{\mathbb{R}^n}}(\mathbb{B}^n) = \overline{\mathbb{B}^n}$ so $\operatorname{Cl_Y}(\operatorname{Int_Y}(D)) = D$

Then by our above arguments we have $\Phi[D] = \operatorname{Cl_X}(e)$ as desired. $\square$

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 $\mathbb{B}^n$.

The answer in this question (Characteristic map of a n-cell in a CW complex) is not applicable here, as in that answer they assume that a chacacteristic map for an open cell is from $\overline{\mathbb{B}^n}$ into $X$ and not from a topological space homeomorphic to $\overline{\mathbb{B}^n}$ as in my question.

Also I'm not entirely sure that the lemma is true (I haven't even proved it), but that was the only way I could get this proof to work. I am sure there must be an easier way to prove this.

Could someone provide a rigorous proof of this?

  • $\begingroup$ Also please indicate why we don't need to bring $Y$ into the picture $\endgroup$ – Perturbative Feb 21 '18 at 17:46
  • $\begingroup$ Your proof is already correct and complete if you just replace $Int_Y(D)$ everywhere by the interior of $D$ as defined in my answer. Your discussion of the interior with respect to $Y$ and $D$ is completely irrelevant because, as I said in my answer, that's not how "the interior of $D$" is defined. $\endgroup$ – Eric Wofsey Feb 21 '18 at 21:53

There is a definition you're missing (in the middle of page 129). If $D$ is an $n$-cell, then the "interior" of $D$ is defined to be the image of $\mathbb{B}^n$ under some homeomorphism $\overline{\mathbb{B}^n}\to D$. It turns out that this definition is independent of the homeomorphism chosen, but you do not need this fact for any of your arguments (you can just pick such a homeomorphism once and for all and use that to define the interior of $D$).

(Incidentally, your lemma is not true. For instance, let $M$ be any space with a subset $A$ that is open but not closed and let $B=N=A$.)


Proof: Let $(X, \Gamma)$ be a cell complex. Pick $e \in \Gamma$ and let $\Phi : D \subseteq Y \to X$ be a characteristic map for $e$. We need to show $\Phi[D] = \operatorname{Cl}_X(e)$

By definition of a characteristic map we have $\Phi$ to be continuous and $\Phi[\operatorname{Int}_Y(D)] =e$. Since $D \subseteq Y$ is a closed $n$-cell by definition there exists a homeomorphism $\psi : \overline{\mathbb{B}^n} \to D$. By bijectivity of $\psi$ we have $\psi[\overline{\mathbb{B}^n}] = D$. Recall that $\operatorname{Int}_Y(D)$ is defined to be $\psi[\mathbb{B}^n]$ so $\psi[\mathbb{B}^n]=\operatorname{Int}_Y(D)$

Now observe that since $\psi$ is a homeomorphism from $\overline{\mathbb{B}^n}$ to $D$ and $\mathbb{B}^n \subseteq \overline{\mathbb{B}^n}$ we have $\psi\left[\operatorname{Cl}_{\overline{\mathbb{B}^n}}(\mathbb{B}^n)\right] = \operatorname{Cl}_Y\left(\psi[\mathbb{B}^n]\right) = \operatorname{Cl}_Y\left(\operatorname{Int}_Y(D)\right)$. Now since $D = \psi[\overline{\mathbb{B}^n}] =\psi\left[\operatorname{Cl}_{\overline{\mathbb{B}^n}}(\mathbb{B}^n)\right]$ we have $\operatorname{Cl}_Y\left(\operatorname{Int}_Y(D)\right) = D$.

With everything set up we now show that $\Phi[D] \subseteq \operatorname{Cl}_X(e)$. By continuity of $\Phi$ we have $\Phi[D] = \Phi[\operatorname{Cl}_Y(\operatorname{Int}_Y(D))] \subseteq \operatorname{Cl}_X(\Phi(\operatorname{Int}_Y(D))) = \operatorname{Cl}_X(e)$

Conversely we now show that $\operatorname{Cl}_X(e) \subseteq \Phi[D]$. Observe that $D$ is compact since it is homeomorphic to the compact space $\overline{\mathbb{B}^n}$ and since $\Phi$ is continuous $\Phi[D]$ is compact.

Moreover since $X$ is Hausdorff we have $\Phi[D]\subseteq X$ is also Hausdorff hence $\Phi$ is a continuous map from a compact space to a Hausdorff space ans is thus closed by the closed map lemma.

Now since $\Phi$ is a closed map we have $\operatorname{Cl}_X(\Phi(\operatorname{Int}_Y(D))) \subseteq \Phi(\operatorname{Cl}_Y(\operatorname{Int}_Y(D))) = \Phi[D]$ and since $\Phi(\operatorname{Int}_Y(D)) = e$ we have $\operatorname{Cl}_X(e) \subseteq \Phi[D]$ and thus we finally arrive at $\Phi[D] = \operatorname{Cl}_X(e)$ as desired. $\square$

I hope others out there find this useful, this took me quite a while to prove as I was looking at it the wrong way. Thanks to @EricWofsey for pointing out that I had missed a definition.


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