I have a question about partitions of unity specifically in the book Calculus on Manifolds by Spivak. In case 1 for the proof of existence of partition of unity, why is there a need for the function $f$? The set $\Phi = \{\varphi_1, \dotsc, \varphi_n\}$ looks like is already the desired partition of unity. Following is the theorem and proof. Only Case 1 in the proof is relevant.
Best Answer
I was looking back at my question today for some reason and immediately saw why the function $f$ is required. Although $\psi_i$ is smooth with compact support in $U_i$, the functions $\varphi_i$ can only be defined on $U$ where $\sum_{i = 1}^n\psi_i > 0$. The problem is that $\varphi_i$ usually does not go to zero at the boundary of $\operatorname{supp}(\psi_i)$ (much less smoothly extend to zero outside the boundary). You can see this near the boundary of a $\operatorname{supp}(\psi_i)$ which is away from all other $\operatorname{supp}(\psi_j)$. Near this boundary $\varphi_i(x) = \frac{\psi_i(x)}{\psi_i(x)} = 1$. The solution is to use a cutoff function $f$ which forces everything to smoothly go to $0$ near the boundary of $U$.