# Proof of the existence of a partition of unity

I am trying to understand the proof that partitions of unity on a smooth manifold exist. To this end, I would like to post the proof in my own words here and kindly request that someone read it and tell me if it is correct. Thank you in advance for helping me learn.

First I would like to give some definitions.

A partition of unity on a $C^\infty$-$n$-manifold $M$ is a collection of smooth functions $\varphi_i : M \to \mathbb R$ such that $\varphi_i \ge 0$ and for all $m \in M$: $\sum_i \varphi_i(m) = 1$ and for only finitely many $i$: $\varphi_i (m) \neq 0$. We assume that a manifold is a second countable locally compact Hausdorff space and we use (without proof) that we can write such a space as a union of sets $G_i$ with the property that $\bigcup_i G_i = M$, $\overline{G_i}$ is compact and $\overline{G_i}\subseteq G_{i+1}$.

I will also use that there exists a smooth map $\varphi: \mathbb R^n \to \mathbb R$ such that $\varphi$ is $1$ on the open cube of radius $1$ and $0$ outside the closed cube of radius $2$.

Now the claim is:

If $M$ is a smooth $n$-manifold and $U_i$ is an open cover of $M$ then there exists a partition of unity with compact support and $\mathrm{supp}(\varphi_i)$ contained in $U_i$.

Here is the proof (how I understand it): Let $U_i$ be a given open cover of the smooth manifold $M$. Pick a point $m \in M$. Then $m \in U_i$ for some $i$. Let $(\psi, V)$ be any chart with $m \in V$. Then since scaling of a chart, translation of a chart and restriction of a chart to an open subset of $V$ yields again a chart we may assume that $\psi (m) = 0$ and that $\psi(V)$ contains the closed cube of radius $2$ and that $V \subseteq U_i \cap G_j \setminus \overline{G_{j-1}}$ where $j$ is the largest $j$ such that $m \notin \overline{G_j}$.

Let $\widetilde{\varphi}: \mathbb R^n \to \mathbb R$ denote that map that is $1$ inside the open cube of radius $1$ centred at $0$ and $0$ outside the cube of radius $2$. Then we define a map

$$\varphi (m) = \begin{cases} 0 & m \notin V \\ \widetilde{\varphi} \circ \psi (m) & m \in V\end{cases}$$

Then it is clear that $\varphi$ is a smooth map $M \to \mathbb R$. Furthermore, since its support is contained in $\overline{V}$ and $\overline{V} \subseteq \overline{G_{j+1}}$ it is clear that the support is compact.

For $m \in \overline{G_i} \setminus G_{i-1}$ let $N_m$ denote an open neighbourhood of $m$ contained in $\overline{G_i} \setminus G_{i-1}$ such that $\varphi$. By construction the set $\overline{G_i} \setminus G_{i-1}$ is compact. For $m \in \overline{G_i} \setminus G_{i-1}$ let $U_m$ denote an open neighbourhood on which the previously constructed map $\varphi$ is $1$. Then since $\overline{G_i} \setminus G_{i-1}$ is compact we may cover it with finitely many such neighbourhoods. We do this for every $i$ so as to obtain a sequence of smooth maps $\varphi_k$. Since the sum $\sum_k \varphi_k$ has only finitely many terms not equal to zero this sum converges at every point in $M$. Hence we may define

$$\varphi_k' = {\varphi_k \over \sum_k \varphi_k'}$$

It is clear that $\varphi_k'$ form a partition of unity for $M$ subordinate to the cover $U_i$.