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.