# The Unapologetic Mathematician

## Partitions of Unity (proof)

Finally we can prove what we’ve asserted: given any open cover $\{U_\alpha\}$ of a smooth manifold $M$ we can find a countable smooth partition of unity $\{\phi_k\}$ subordinate to it.

So, as we’ve seen we can find a countable atlas $(V_k,x_k)$; we use $x_k$ for the coordinate maps since we’ll want the $\phi_k$ free. We’ve also seen that we have a smooth bump function $\phi$ between the two cubes $C_1(0)$ and $C_2(0)$ in $\mathbb{R}^n$. So let’s define

$\displaystyle\theta_k(p)=\left\{\begin{array}{lc}\phi\left(x_k(p)\right)&p\in V_k\\{0}&p\notin V_k\end{array}\right.$

Now, it’s easily verified that the furthest points from the origin in $C_2(0)$ are the corners, each of which is $2\sqrt{n}$ away. Thus we can tell that $C_2(0)\subseteq B_{3n}(0)$, and the support of $\phi$ is contained entirely within $x_k(V_k)=B_{3n}(0)$. This means that the support of $\theta_k$ is entirely contained within $V_k$ — by the time we get to the edge it’s already smoothly tailed off to zero, and so even though we define it piecewise, $\theta_k$ is a smooth function defined on all of $M$.

Now set up the sum

$\displaystyle\theta(p)=\sum\limits_k\theta_k(p)$

Since $V_k$ is locally finite and $\theta_k$ is supported within $V_k$, this sum is guaranteed to be finite at each point, which makes $\theta$ a smooth function on all of $M$. The ball $B_1(0)$ is contained within the cube $C_1(0)$, so $\phi$ takes the constant value $1$ on this ball. Since the preimages $x_k^{-1}(B_1(0))$ form an open cover of $M$, there is always at least one $k$ for which $\theta_k=1$. In particular, for which it’s not zero, and thus the whole sum is nonzero at $p$.

Since $\theta$ is never zero, we can divide by it. We define $\phi_k(p)=\theta_k(p)/\theta(p)$. These are smooth functions on all of $M$, and their sum is everywhere exactly $1$. Thus the $\phi_k$ form a partition of unity subordinate to $\{V_k\}$. And since $\{V_k\}$ refines $\{U_\alpha\}$, the partition is subordinate to this cover as well.