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Let´s say we have a partition of unity $\{\chi\}_{i \in I}$ subordinate to $\{U_i\}_{i \in I}$ where $U_i$ can be found in the atlas $$\mathcal{A}_{M} := \{(U_i,\varphi_i):i \in I\}$$ on a smooth manifold $M$ with boundary $\partial M$. We get an induced atlas $$\mathcal{A}_{\partial M} := \{(U_i \cap \partial M, \varphi_i|_{U_i \cap \partial M}):i \in I\}$$ on $\partial M$. Now, it is not entirely obvious to me that $\{\chi\}_{i \in I}$ will be subordinate to $\{U_i \cap \partial M\}_{i \in I}$. It is clear that $$\sum_{i \in I} \chi_i \equiv 1$$ and it is also clear that for every $p \in \partial M$, there exists a neighbourhood $V$ such that $$\operatorname{supp}\chi_i \cap V \neq \emptyset$$ for only finitely many $i \in I$, since it holds for $\{U_i\}_{i \in I}$ (I suppose you can just take $V \cap \partial M$). It is however not clear that $\operatorname{supp} \chi_i \subset U_i \cap \partial M$. But it seems like my notes assume (without proof) that this is true, in the proof of the generalized stokes theorem. I am however not sure, it is not said explicitly that this is the case. Is it true that the "subordination carries over" to the induced atlas on the boundary, or can one make amends to the original family of non-negative functions $\{\chi_i\}_{i \in I}$ to fix this?

Ben123
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In general for any topological space $M$, any function $f:M\to\Bbb{R}$, and any subset $E\subset M$, we have the inclusion \begin{align} \text{supp}\left(f|_E\right)\subset \text{supp}(f)\cap E. \end{align} Keep in mind that $\text{supp}(f|_E)$ is defined as the closure in $E$ of the set \begin{align} \{x\in E\,:\, f|_E(x)\neq 0\}=\{x\in E\,:\, f(x)\neq 0\}. \end{align} I omit the proof since it is easy. (For completeness, I’ll mention that the reverse inclusion can fail spectacularly, as it is possible for $\text{supp}(f|_E)=\emptyset$ and $\text{supp}(f)=M$).


In your question, to be super clear, they should refer to the restrictions of the $\chi_i$, i.e $\chi_i|_{\partial M}$ when talking about the induced partition of unity.

peek-a-boo
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