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Jul 7 at 15:01 vote accept CommunityBot
Jul 7 at 13:31 comment added Z. M @user531303 Sorry, I am unable to understand most of your comments, but here are some remarks: 1. if you take termwise global sections to a quasi-isomorphism of cochain complexes, the map you get might not be a quasi-isomorphism, and the paracompactness is used to guarantee this; 2. the sheaves are defined in the usual way, i.e., on the open subsets $\operatorname{Op}(M)$ of $M$, and $\operatorname{Op}(M)$ is not a full subcategory of manifolds; 3. I am not sure what you want to say about effective descent.
Jul 7 at 13:28 history edited Z. M CC BY-SA 4.0
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Jul 7 at 0:26 comment added user531303 Ok, after some thought, maybe I see: the existence of a countable atlas (same as paracompact), gives that one can divide the above into the case of a finite atlas and the case of a countable-concentric atlas $U_1 \subset U_2 \subset \cdots ...$. In each case, the tensor product commutes with the product. Separately, $C^{\infty}(U_x) \otimes_{\mathbb{R}} C^{\infty}(U_y) \cong C^{\infty}(U_x \cap U_y)$.
Jul 6 at 23:30 comment added user531303 Suppose that the manifold is locally diffeomorphic to $\mathbb{R}^n$ (unlike, for example, the open square with dictionary order), and pick an open affine neighborhood $U_x$ of $x$ for each point $x$ in the smooth manifold $M$. In this case, $C^{\infty}(M) \rightarrow \Pi_{x \in M} C^{\infty}(U_x)$ is effective descent (faithfully flat), and $C^{\infty}(M)$ is isomorphic to the kernel of $\Pi_{x \in M} C^{\infty}(U_x) \implies \Pi_{x, y \in M} C^{\infty}(U_x \cap U_y)$.
Jul 6 at 21:09 comment added user531303 Oh, I see- the domain of the sheaf is a fully faithful subcategory of smooth manifolds.
Jul 6 at 21:06 comment added user531303 Sorry, I had misunderstood the role of the sheaves here. Is it no different to take global sections in 1 given the local compactness condition?
Jul 6 at 19:00 history edited Z. M CC BY-SA 4.0
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Jul 6 at 14:41 history answered Z. M CC BY-SA 4.0