I'm trying to prove Lemma 8.2. from this lecture note.
Lemma 8.2. Let $M$ be an $m$-manifold and $W$ an open set in $M$. Let $x \in W$. Then there is a smooth function $\theta: M \longrightarrow[0,1]$ such that $\theta =0$ on $M \setminus W$ and that $\theta = 1$ on some neighbourhood of $x$.
Could you verify if my understanding is fine?
Proof. We need the following result, i.e.,
Lemma 8.1. There is a smooth function, $\theta_0: \mathbb{R}^m \longrightarrow[0,1] \subseteq \mathbb{R}$, with $\theta_0(x)=1$ whenever $\|x\| \leq 1$, and with $\theta_0(x)=0$ whenever $\|x\| \geq 2$.
Let $\phi: U \longrightarrow V \subseteq \mathbb{R}^m$ be a chart with $x \in U$. Let $\theta_0$ be given by Lemma 8.1. By translation and dilation, we can suppose that $0 \in \phi(W \cap U)$ and $B(0, 3) \subset \phi(W \cap U)$. Now set $\theta(x)=\theta_0(\phi(x))$ for $x \in U$ and set $\theta(x)=0$ for $x \in M \backslash U$. Let $\psi:A \to B \subset \mathbb R^m$ be a chart. We need to show that $$ \theta \circ \psi^{-1} :B \to \mathbb R^m $$ is smooth. Fix $x \in B$ and let $y :=\psi^{-1} (x)$.
- Assume $y \in U$. Then $\theta$ is smooth at $x$ by chain rule.
- Assume $y \notin U$. It suffices to show that there is an open neighborhood $O$ of $y$ such that $\theta_0 \circ \phi =0$ on $O$. We have $\theta_0 \circ \phi =0$ on $C :=M \setminus \phi^{-1} ( \overline{B(0, 2)} )$. Clearly, $C$ is open in $M$. Because $y \notin U$ and $B(0, 3) \subset \phi(U)$, we get $$ y \in M \setminus \phi^{-1} (B(0, 3)) \subset C. $$ Then we can pick $O$ such that $y \in O \subset C$.