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I am following Lee in his introduction to smooth manifolds page 442. He explains that if $\{M_j\}_{j = 1}^\infty$ is a countable collection of smooth $n$-manifolds with or without boundary and let $M = \coprod_{j=1}^\infty M_j$, then there is an isomorphism of the de Rham cohomology groups $H^p(M)$ and $\prod_{j=1}^\infty H^p(M_j)$.

But what is the standard smooth structure that turns $\coprod_{j=1}^\infty M_j$ into a smooth manifold?

Mikkel Rev
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    I think $C^\infty \left (\coprod_i M_i \right)$ is simply $\prod_i C^\infty(M_i)$. – isekaijin Jun 01 '19 at 20:39
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    Every open subset of $\coprod_i M_i$ is of the form $\coprod_i U_i$ with $U_i$ open in $M_i$, so you can reconstruct the entire sheaf of smooth functions from that. – isekaijin Jun 01 '19 at 20:41
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    @pyon How about taking the disjoint union of the smooth structures of each $M_j$, $\mathcal{A} = \coprod_j \mathcal{A}_j$? Do you think this would make a good smooth structure? – Mikkel Rev Jun 01 '19 at 20:48
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    A smooth structure on $M$ is first and foremost a presheaf, i.e., a contravariant functor from the topology of $M$ (as a lattice, i.e., as a particular small category) that sends each open $U \subset M$ to the ring of smooth functions $C^\infty(U)$. This satisfies a couple of gluing conditions making it, in fact, a sheaf. – isekaijin Jun 01 '19 at 21:20
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    It is not clear to me what a disjoint union of functors should be, though. – isekaijin Jun 01 '19 at 21:22
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    Yes, $\coprod_j \mathcal A_j$ is what you want. – Lee Mosher Jun 02 '19 at 14:37
  • @LeeMosher Sorry if I sound stupid. But do you mean the disjoint union of atlases? – Adittya Chaudhuri Feb 27 '20 at 17:23

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