> [!theorem] Lee Theorem 2.23 > > Let $X$ be a [[n-Manifold|smooth n-manifold]] and $\seqi{U}$ be an [[Open Cover|open cover]], then there exists a smooth partition of unity subordinate to it. > > *Proof*. Let $\bracs{B_{i, n}}$ be a countable base for $U_i$ consisting of regular coordinate balls, then $\bracs{B_{i, n}: i \in I, n \in \nat}$ is a base for $X$ consisting of regular coordinate balls. Since $X$ is a topological manifold, there exists a countable, locally finite refinement $\bracs{B_j}_{j \in J}$. > > For each $j$, there exists a chart $(U, \varphi)$ and an open ball ball $\widehat{B_j} \subset \widehat{U}$ such that $\widehat{B_j} = \varphi(B_j)$. By translating $\varphi$, assume without loss of generality that $\widehat{B_j}$ is centred at $0$. If $B_j \subset U_i$, then since it is a regular coordinate ball in $U_i$, there exists a coordinate ball $B_j \subset U_i$ such that such that $B_j' \supset \ol{B_j}$. Let $H_j: \widehat{U} \to \real$ be the smooth bump function that is positive on $\ol{B_j}$, and zero elsewhere, and let > $ > f_j(x) = \begin{cases} > H_j \circ \varphi(x) &x \in B_j' \\ > 0 &x \in M \setminus \ol{B_j} > \end{cases} > $ > which is well-defined as $H_j \circ \varphi(x) = 0$ for all $x \not\in \ol{B_j}$. Since $f_j$ is smooth in $\ol{B_j}$ by construction of $H_j$, and constant elsewhere, $f_j$ is smooth. Moreover, the support of $f_j$ is contained in $\ol{B_j} \subset U_i$. > > Let $f = \sum_{j \in J}f_j$, then since $\seqj{B}$ cover $X$, $f(x) > 0$ for any $x$. Let > $ > F_j = \frac{f_j}{\sum_{k \in J}f_k} > $ > then as $\seqj{B}$ is locally finite, the bottom sum is finite, and $F_j$ is smooth at each point. From here, $\bracs{F_j}_{j \in J}$ is a smooth partition of unity subordinate to $\seqi{U}$.