A chart bump at a point with prescribed support

If $M$ has dimension zero, each singleton is open by the chart condition and is closed since $M$ is Hausdorff. Define $\rho (p)=1$ and $\rho =0$ elsewhere. This is locally constant, hence smooth in charts, and its support is $\{p\}\subseteq W$. Henceforth assume the dimension $n$ is positive.

Choose a smooth chart $(U,\phi )$ with $p\in U$ and put $a=\phi (p)$. Since $\phi (W\cap U)$ is open, choose $\epsilon >0$ with $B_{\epsilon }(a)\subseteq \phi (W\cap U)$, and put $R=\epsilon \mathrm{/}2$, $r=R\mathrm{/}2$. Then ${\bar{B}}_R(a)\subseteq \phi (W\cap U)$. By [L3], ${\bar{B}}_r(a)$ is compact. Apply [L1] inside $B_R(a)$ to obtain a smooth $\tilde{\rho }:{\mathbb{R}}^n\to [0,1]$ equal to one on ${\bar{B}}_r(a)$ with support in $B_R(a)$.

The closed ball ${\bar{B}}_R(a)$ is compact by [L3]. Its image $K={\phi }^{-1}({\bar{B}}_R(a))$ under the continuous inverse chart, viewed as a map into $M$, is compact by [L4]. Thus $K$ is closed in $M$ and $K\subseteq W\cap U$.

On the open cover $U\cup (M\setminus K)$, use $\rho =\tilde{\rho }\circ \phi$ on $U$ and $\rho =0$ on $M\setminus K$. The functions agree on $U\setminus K$, since its coordinate image misses ${\bar{B}}_R(a)$ and hence the support of $\tilde{\rho }$. Composition and pasting in [L2] give a smooth global real-valued function whose values lie in $[0,1]$.

The function equals one at $p$ by step 2.1. Its nonzero locus is contained in the closed set $K$ by step 4.1, so [F2] gives $\mathrm{supp}\rho \subseteq K\subseteq W$. This proves the positive-dimensional case, and step 1.1 proves the zero-dimensional case.