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such that $\lambda_{\alpha} \cdot g_{i} \neq 0$ only for $i \in I_{\alpha}$. Also note that $\lambda_{\alpha} \cdot g_{i} \in \mathcal{J}_{N}$. Then $\sum_{i \in I} (g_{i} \circ \psi) h_{i} = \sum_{i \in I} ((\sum_{\alpha=0}^{m}\lambda_{\alpha}) \cdot g_{i} \circ \psi) h_{i} = (\sum_{i \in I} (g_{i} \circ \psi) h_{i}) \cdot (\lambda_{0} \circ \psi)$ $+ \sum_{\alpha=1}^{m} \sum_{i \in I_{\alpha}} (\lambda_{\alpha} \cdot g_{i} \circ \psi) h_{i}$. The first summand is in $\mathcal{I}$ as $\lambda_{0} \in \mathcal{J}_{N}$ and the rest is a finite sum of elements in $\mathcal{I}$.
Version 2 can be significantly simplified. By the modification of Lemma 1 above, you can assume that $f_{i} = (g_{i} \circ \psi) h_{i}$, where $\{ supp(g_{i}) \}_{i \in I}$ is locally finite. It then suffices to cover $N$ by finitely many precompact sets $\{U_{\alpha} \}_{\alpha=1}^{m}$ (e.g. coordinate balls). Let $U_{0} := M - N$ and let $\{ \lambda_{\alpha} \}_{\alpha = 0}^{m}$ be the corresponding partition of unity. Note that $\lambda_{0} \in \mathcal{J}_{N}$. Now, as $U_{\alpha}$ are compact for $\alpha > 0$, there is a finite subset $I_{\alpha} \subseteq I$ such that
Your answer related to the original formulation of this question (I have edited it significantly). In a summary - you give an answer for the case where all the manifolds are ordinary manifolds. However, in supermanifolds, one does not have a full apparautus of tubular neighborhoods, or the fact that k-dimensional manifold can be covered by k+1 charts.
Thank you. In the end, I have kind of combined your approach with the one in the lecture notes I have posted (I mostly tried to avoid tubular neighborhoods as I am not completely unfailing in their understanding :)). I learned about compact exhaustions along the way (nice thing!). This is where I have to stop and return to the original paper :) Thank you for your help.
Yeah, I also think it is not important, that is why added this separately. However, it is entirely possible that the statement holds only for the particular $\mathcal{V}$ they are considering (although I don't think so), so it does not hold in full generality...
That is very nice, thank you! I understand why you need compactness in your proof, and I see not way around this at first glance. Unfortunately, I need it without this restriction - but I will definitely take note of your approach, thanks.
