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I finished reading Lee's 'introduction to topological manifolds' (2nd edition) and I'm currently tying up some loose ends. One thing I can't understand is the proof of paracompactness of CW complexes. The proof contains some mistakes I feel (perhaps wrongly), and although I get the general idea, there is one technical detail I can't work out. Lee uses a somewhat different approach (as far as I know), just to illustrate a technique of inductively building maps out of a CW complex skeleton par skeleton. He claims that any open cover $\left ( U_\alpha \right )_{\alpha \in A}$ of a CW complex $X$ has a partition unity $\left ( \psi_\alpha \right )_{\alpha \in A}$ subordinate to it, from which paracompactness follows (I guess because then $\left ( \psi_\alpha^{-1}\left \{ x \in X| \psi_{\alpha }(x)\neq 0 \right \}\right )_{\alpha \in A}$ is a locally finite open refinement of $\left ( U_\alpha \right )_{\alpha \in A}$). The aim is to inductively build a partition of unity $\left ( \psi^n_\alpha \right )_{\alpha \in A}$ of the n-skeleton $X_n$ subordinate to the open cover $\left ( U^n_\alpha \right )_{\alpha \in A}$ of $X_n$, where $U^n_\alpha = U_\alpha \cap X_n$, as follows: for $n=0$, choose, if $x \in X_0$, any $\alpha(x) \in A$ such that $x \in U_{\alpha(x)}$. Then, set $\psi^0_{\alpha(x)}(x)= 1$ and $\psi^0_{\alpha}(x)= 0$ if $\alpha \neq \alpha(x)$. Then, suppose that we have found, for $k=0,\dots,n$, partitions of unity $\left ( \psi^k_\alpha \right )_{\alpha \in A}$ of $X_k$ subordinate to $\left ( U^k_\alpha \right )_{\alpha \in A}$, such that $ {\psi^k_\alpha}_{|X_{k-1} }=\psi^{k-1}_\alpha$ and, if $\psi^{k-1}_\alpha\equiv 0$ on an open subset $V$ of $X_{k-1}$, then there exists an open subset $V'$ of $X_k$ containing $V$ on which $\psi^{k}_\alpha \equiv 0$. Let $q:X_n\sqcup \bigsqcup_{\gamma \in \Gamma} D^{n+1}_{\gamma} \rightarrow X_{n+1}$be the quotient map realizing $X_{n+1}$ as an adjunction space obtained by attaching $n+1$- cells $D^{n+1}_{\gamma}$ to $X_n$ and let $\phi_\gamma : \partial D^{n+1}_{\gamma} \rightarrow X_n$ be the maps glueing $ \partial D^{n+1}_{\gamma}$ to $X_n$. First, set $\tilde{\psi}^n_{\alpha,\gamma}=\psi^n_\alpha \circ \phi_\gamma :\partial D^{n+1}_{\gamma} \rightarrow \left [ 0,1 \right ]$ and $\tilde{U}^{n+1}_{\alpha,\gamma}=q_{|D^{n+1}_\gamma}^{-1}(U^{n+1}_\alpha)$, so, in particular, $(\tilde{U}^{n+1}_{\alpha,\gamma})_{\alpha \in A}$ is an open cover of $D^{n+1}_\gamma$ and choose a partition of unity $(\eta_{\alpha,\gamma})_{\alpha \in A}$ of $D^{n+1}_\gamma$ subordinate to $(\tilde{U}^{n+1}_{\alpha,\gamma})_{\alpha \in A}$. For $\gamma \in \Gamma$ fixed, $(supp \ \psi^n_\alpha)_{\alpha \in A}$ is a locally finite family of subsets of $X_n$ by construction and since $\phi_\gamma(\partial D^{n+1}_{\gamma})$ is compact in $X_n$, it meets $supp \ \psi^n_\alpha$ for at most a finite amount of indices (I get that). Hence, only for finitely many indices $\alpha_1,\dots,\alpha_k$, we have $\tilde{\psi}^n_\alpha \not\equiv 0$ (I get that). Let, for $C\subseteq \partial D^{n+1}_{\gamma} $, and $0<\varepsilon <1$, $C(\varepsilon)=\left \{ x \in D^{n+1}_{\gamma} |\frac{x}{\left \| x \right \|} \in C \text{ and } 1-\varepsilon < \left \| x \right \| \leqslant 1 \right \}$, norm confer some homeomorphism from $ D^{n+1}_{\gamma}$ to the closed unit ball. For any $\alpha_j$ of the indices $\alpha_1,\dots,\alpha_k$, $supp \ \tilde{\psi}^n_{\alpha_j,\gamma}$ is a non-empty compact subset of $\partial D^{n+1}_{\gamma}$ so of $D^{n+1}_{\gamma}$ and is contained in $\tilde{U}^{n+1}_{\alpha_j,\gamma}$ (I can work that out). Since $co \tilde{U}^{n+1}_{\alpha}$ is closed in $D^{n+1}_{\gamma}$ and disjoint from $supp \ \tilde{\psi}^n_{\alpha_j,\gamma}$ , one can find an $0<\varepsilon_j<1$ such that $supp \ \tilde{\psi}^n_{\alpha_j,\gamma} (\varepsilon_j)\subseteq \tilde{U}^{n+1}_{\alpha_j,\gamma}$ (I get that). Then, set $\varepsilon_\gamma =\min^k_{j=1} \varepsilon_j$ (I would feel more confident choosing $0<\varepsilon_\gamma <\min^k_{j=1} \varepsilon_j$ ) and let $\sigma: D^{n+1}_\gamma \rightarrow \left [0,1 \right ]$ be a bump function that is $1$ on $D^{n+1}_\gamma \setminus \partial D^{n+1}_\gamma(\varepsilon_\gamma) $ and is supported in $\partial D^{n+1}_\gamma(\dfrac{\varepsilon_\gamma}{2})$. This seems wrong to me because $\partial D^{n+1}_\gamma(\dfrac{\varepsilon_\gamma}{2})$ and $D^{n+1}_\gamma \setminus \partial D^{n+1}_\gamma(\varepsilon_\gamma) $ are disjoint. I think that $\partial D^{n+1}_\gamma(\dfrac{\varepsilon_\gamma}{2})$ should be replaced with something like $D^{n+1}_\gamma \setminus \overline{\partial D}^{n+1}_\gamma(\dfrac{\varepsilon_\gamma}{2}) $ . So, let's do that. Then define $\tilde\psi ^{n+1}_{\alpha,\gamma}: D^{n+1}_\gamma \rightarrow \left [ 0,1 \right ]$ by $\tilde\psi ^{n+1}_{\alpha,\gamma}(x)=\sigma(x)\eta _{\alpha,\gamma}(x)+(1-\sigma(x))\tilde{\psi}^n_{\alpha,\gamma}(\frac{x}{\left \| x \right \|}) $. $\tilde\psi ^{n+1}_{\alpha,\gamma}$ is continuous, takes values in $\left [ 0,1 \right ]$, coincides with $\tilde\psi ^{n}_{\alpha,\gamma}$ on $\partial D^{n+1}_\gamma$ and $\sum_{\alpha \in A}\tilde\psi^{n+1}_{\alpha,\gamma}\equiv 1$. $supp \ \tilde{\psi}^{n+1}_{\alpha,\gamma}\subseteq \tilde{U}^{n+1}_{\alpha,\gamma}$ (I can work that out, but only if one chooses $\varepsilon_\gamma< \min^{k}_{j=1}\varepsilon_j$ ). Now, repeat this for every $\gamma \in \Gamma$. The map that coincides with $\psi^n_\alpha$ on $X_n$ and with $\tilde{\psi}^{n+1}_{\alpha,\gamma}$ on $D^{n+1}_\gamma$ passes to the quotient to yield the requested $\psi^{n+1}_\alpha$. Okay, here is my problem: I cannot relate the supports of $\psi^{n}_\alpha$ and the $\tilde{\psi}^{n+1}_{\alpha,\gamma}$ to the support of $\psi^{n+1}_\alpha$ accurately enough. I can only show that $supp \ \psi^{n+1}_\alpha \subseteq \overline{U}^{n+1}_\alpha$ . Any thoughts anyone?

I can understand the rest of the proof, however, something seems not right, so, to be sure, here it is. To check that $(supp \ \psi^{n+1}_\alpha)_{\alpha \in A}$ is locally finite, suppose $x$ is in the interior of an $n+1$-cell $D^{n+1}_\gamma$. Lee claims that the cell itself is a neighbourhood of $x$ on which only finitely many of the $\psi^{n+1}_\alpha$ are non-zero. However, is this true? Identically zero $\tilde{\psi}^{n}_\alpha$ cán yield non-zero $\tilde{\psi}^{n+1}_\alpha$. Having said that, there is an easy way to work around this, because of the use of the $\eta_{\alpha,\gamma}$ in the construction of the $\tilde{\psi}^{n+1}_\alpha$. Any comments would be grately appreciated, big thanks in advance.

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+1 for the effort –  Ilya Jan 31 '13 at 19:13

2 Answers 2

I commend you for your careful reading! You're right on two counts.

Then, set $\varepsilon_\gamma =\min_{j=1}^k\varepsilon_j$ (I would feel more confident choosing $0<\varepsilon_γ<\min_{j=1}^k\varepsilon_j$)

You're absolutely right. This is needed in order to ensure that the function $\tilde\psi^{n+1}_{\alpha,\gamma}$ (in your notation) is supported in $\tilde{U}^{n+1}_{\alpha}$. I've added a correction to my online correction list.

I think that $\partial D^{n+1}_\gamma(\varepsilon/2)$ should be replaced with something like $D^{n+1}_\gamma(\varepsilon/2)\setminus \partial D^{n+1}_\gamma(\varepsilon/2)$

Also correct. I've also added this to my corrections.

I cannot relate the supports of $\psi^{n}_\alpha$ and the $\tilde{\psi}^{n+1}_{\alpha,\gamma}$ to the support of $\psi^{n+1}_\alpha$ accurately enough. I can only show that $supp \psi^{n+1}_\alpha \subseteq \overline{U}^{n+1}_\alpha$ .

With the corrections noted above, I think this works. Let's see if I can explain it a little more convincingly. Suppose $x$ is a point of $X_{n+1}$ that is not in $U^{n+1}_\alpha$; we need to show that $\psi^{n+1}_\alpha$ is zero on a neighborhood of $x$. If $x\in X_n$, then it has a neighborhood $V\subseteq X_n$ on which $\psi^{n}_\alpha\equiv 0$. If $x$ doesn't meet the closures of any $n+1$-cells, then $V$ is also a neighborhood of $x$ in $X_{n+1}$. Otherwise, for each $\gamma$ such that $x\in \phi_\gamma(\partial D^{n+1}_\gamma)$, by construction $V(\varepsilon/2)$ is an open subset of $ D^{n+1}_\gamma$ on which $\tilde\psi^{n+1}_{\alpha,\gamma}\equiv 0$. The union of these sets is a saturated open subset of the disjoint union, whose image in $X_{n+1}$ is a neighborhood of $x$ on which $\psi^{n+1}_\alpha\equiv 0$. On the other hand, if $x\in X_{n+1}\setminus X_n$, then it lies in (the image of) the interior of some $D^{n+1}_\gamma$. In that case, it does not lie in $\tilde U^{n+1}_\alpha$, so both terms in the definition of $\tilde \psi^{n+1}_{\alpha,\gamma}$ are zero on a neighborhood of $x$ in $\mathrm{Int}\ D^{n+1}_\gamma$, which is also a neighborhood of $x$ in $X_{n+1}$.

To check that $(supp \ \psi^{n+1}_\alpha)_{\alpha \in A}$ is locally finite, suppose $x$ is in the interior of an $n+1$-cell $D^{n+1}_\gamma$. Lee claims that the cell itself is a neighbourhood of $x$ on which only finitely many of the $\psi^{n+1}_\alpha$ are non-zero. However, is this true? Identically zero $\tilde{\psi}^{n}_\alpha$ cán yield non-zero $\tilde{\psi}^{n+1}_\alpha$.

It's true that $\tilde{\psi}^{n+1}_\alpha$ can be nonzero on $D^{n+1}_\gamma$ even if $\tilde{\psi}^{n}_\alpha$ is identically zero there. However, because $(\eta_{\alpha,\gamma})_{\alpha \in A}$ is a partition of unity and $D^{n+1}_\gamma$ is compact, there are only finitely many indices $\alpha$ such that $\eta_{\alpha,\gamma}$ is not identically zero.

Does this help? Am I missing something? (Unfortunately, it wouldn't be the first time!)

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Dear Prof. Lee, you should hang out more often on Math.SE! Then beginners like me can learn from the pros :D –  user38268 Feb 7 '13 at 17:48
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This interaction is quite amazing! –  Pedro Tamaroff Feb 7 '13 at 19:15
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I thought this sort of interaction was possible only on MathOverflow. It seems that Math StackExchange is on the right track for progression! –  Haskell Curry Feb 7 '13 at 20:57

I had already given up hope on this, but you have answered all my questions, thank you very much. I badly wanted to fully understand this proof because it sits beautifully in the very nice new section on CW complexes and the style of the proof agrees perfectly with the style of the whole book. I feel there is still one thing 'missing' (in fact, it's all there, but it gets no mention), please forgive me if I'm wrong. When building an open $V' \subseteq X_{n+1}$ on which $\psi^{n+1}_\alpha\equiv 0$ out of an open $V \subseteq X_{n}$ on which $\psi^{n}_\alpha\equiv 0$, one has that $V'\cap X_n = V$ by construction. I feel that this is essential and should be mentioned as a prerequisite in the building process, and here is why: Having found $(\psi^{n}_\alpha)_{\alpha \in A}$, a partition of unity of the $n$-skeleton $X_n$ subordinate to $(U^{n}_\alpha=U_\alpha\cap X_n)_{\alpha \in A}$ for each $n$, $\psi_\alpha$ is (evidently) defined to be the function who's restriction to each $X_n$ is $\psi^n_\alpha$ (by the way, there is no mention there of the fact that $supp \psi_\alpha \subseteq U_\alpha$ which (however) can easily be proved essentially in the same way that $(supp \psi_\alpha )_{\alpha \in A}$ is proved to be locally finite). To prove local finiteness of $(supp \psi_\alpha )_{\alpha \in A}$, if $x \in X$, say $x \in X_n$, so $\psi^n_\alpha \equiv 0$ in a neighbourhood $V_n$ of $x$ in $X_n$ except for finitely many of the $\alpha \in A$ (by the local finiteness of $(supp\psi^{n}_\alpha)_{\alpha \in A}$ ).So there exists an open $V_{n+1}\subseteq X_{n+1}$ on which $\psi^{n+1}_\alpha \equiv 0$ with $V_n \subseteq V_{n+1}$, an open $V_{n+2}\subseteq X_{n+2}$ on which $\psi^{n+2}_\alpha \equiv 0$ with $V_{n+1} \subseteq V_{n+2}$ and so on. Now, the claim is that $V=\bigcup_{k\geqslant n} V_k$ is a neighbourhood of $x$ in $X$ on which $\psi_\alpha \equiv 0$, for all but finitely many of the $\alpha \in A$. I guess that one should use the fact that the topology of $X$ is coherent with the collection of its skeletons to check that $V$ is open $X$. However, without the constraint that $V_{n+1}\cap X_n = V_n$,$V_{n+2}\cap X_{n+1} = V_{n+1}$ and so on, there is no way of identifying $V\cap X_k$ for $k>n$.

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