0
$\begingroup$

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 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?

$\endgroup$
  • $\begingroup$ 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
3
+100
$\begingroup$

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 $B=N=A$.)

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.