de Rham's Theorem using atlases and colimit preservation

Question

$\DeclareMathOperator\DR{DR}\newcommand\SmoothManifold{\mathrm{SmoothManifold}}\newcommand\ChainComplexes{\mathrm{ChainComplexes}}\DeclareMathOperator\Sing{Sing}\newcommand\DifferentialGradedRAlgebras{\mathrm{DifferentialGradedℝAlgebras}}$A smooth manifold is a union of its affine open sub-manifolds. A smooth manifold is a colimit of smooth affine submanifolds, which can be constructed from an atlas.

This is related to the concept of glueing data for functions, in which, given an open cover $\{ U_i \}_{i \in I}$ of a topological space $X$, and a topological space $Y$, there is a correspondence between continuous (smooth, holomorphic, etc.) functions $f$ defined on $X$ and continuous (smooth, holomorphic, etc.) $f_i : U_i \rightarrow Y$ such that $f_i|_{U_i \cap U_j} = f_j|_{U_i \cap U_j}$.

My question is about a proof of de Rham's theorem which seems correct, but which I can't find elsewhere. The idea is that the theorem reduces like so:

[A] Reduction under colimits arising from open covers via affine submanifolds

[B] Reduction under products

[C] Demonstration for the case of $[-1,1]$

In light of C, we may prefer to use smooth manifolds with boundary or to switch $[-1,1]$ with $(-1,1)$.

Under the plan above, here are my five questions:

(1) What is the easiest way to show that $\DR(X \times Y) \cong \DR(X) \otimes_{\mathbb{R}} \DR(Y)$ as a differential graded $\mathbb{R}$-algebra?

(2) Does the singular complex functor $\Sing : \SmoothManifold \rightarrow \ChainComplexes$ send colimits to colimits?

(3) Does the functor from smooth manifolds to $\mathbb{R}$-algebras sending a smooth manifold to the smooth $\mathbb{R}$-valued functions on it send colimits of open sub-manifolds to cofiltered limits of $\mathbb{R}$-algebras?

(4) [MAIN THEOREM] Does the DeRham complex functor $\DR : \SmoothManifold \rightarrow \DifferentialGradedRAlgebras$ send colimits arising from open covers to limits of differential graded $\mathbb{R}$-algebras?

(5) What is the easiest way to show that the map of differential graded $\mathbb{R}$-algebras $\DR([-1,1]) \rightarrow [S([-1,1]),\mathbb{R}]$ is a weak equivalence?

Progress:

1. $X$ and $Y$ are locally compact. Take points $x$ in $X$ and $y$ in $Y$, and $f : C^{\infty}(X \times Y)$, we can consider $f_1 = f((-,y))$ and $f_2 = f((x,-))$. $f$ is determined by a power series in a neighborhood of $(x,y)$, and this power series is determined by the power series for $f_1$ and $f_2$, showing that $C^{\infty}(X \times Y)$ is locally contained in the $\mathbb{R}$-vector space product $C^{\infty}(X) \otimes C^{\infty}(Y)$.

2. This follows from the existence of a right adjoint of a similar functor to the singular complex construction for smooth manifolds- the singular complex as defined on CW-complexes (without base).

It seems that $\mathbb{Z}[[\Delta{}^n,X]]/Z(\mathbb{Z}[[\Delta{}^n,X]])$ commutes with colimits because of how one can decompose $\Delta{}^n$ into smaller $\Delta{}^n$'s.

3. A function is determined by its values on stalks ($\mathbb{R}$-algebras of germs $\mathcal{G}$), so that we can replace $C^{\infty}(X)$ with its image in $\prod_{x \in X} \mathcal{G}_n$, and likewise for $C^\infty (U)$, using an $\mathbb{R}$-algebra injection, and where $n$ is the dimension. $\mathcal{G}_n$ is defined as the germ at $0$ of smooth real valued functions on $\mathbb{R}^n$.

A key lemma goes like this:

Stalk Lemma: For $(f_x)_{x \in X}$ in $\prod_{x \in X} \mathcal{G}_n$, suppose that for each $x \in X$, there is a neighborhood $U$ of $x$ in $X$ and a smooth function $f$ defined on $U$ such that the germ of $f$ at $x$ matches $f_x$. Then there is a smooth function $f$ defined on $X$ which matches $f_x$ at each respective stalk (up to isomorphism of stalks $\mathcal{G}_x \cong \mathcal{G}_n$).

4. This is the main problem I want to focus on.

It seems like the fact that DR(X) is finitely generated over $C^{\infty}(X)$ is key since in any expression (formal expression as in the free DGA construction), one could use the stalk lemma after intersecting finitely many open neighborhoods.

The corresponding statement for the $\mathbb{R}$-linear dual of the singular complex ought to follow from generalities concerning adjunctions.

5. I would like to construct a commutative diagram like so:

One more question: it is typical in De Rham's theorem to use partitions of unity, expressing 1 as a sum of functions which are supported on a smooth affine submanifold. The approach here uses projections instead, and requires fewer assumptions. (edit: ) My question is whether De Rham's theorem holds for locally compact manifolds.

Answer 1

I would not believe that the paracompactness is avoidable. In any case, Dustin Clausen's Lectures on algebraic de Rham cohomology, in particular, Lecture 1, should cover some of ingredients that you want. The argument works as follows:

Let $M$ be a manifold.

1. The map $\underline{\mathbb R^{\operatorname{disc}}}\to(\Omega_{(-)}^\ast,\operatorname d)$ of cochain complexes of sheaves on $M$, where the constant sheaf $\underline{\mathbb R^{\operatorname{disc}}}$ is viewed as a cochain complex concentrated in degree 0, is a quasi-isomorphism. This does not need paracompactness, and it reduces to local charts, which is Poincaré's lemma, and his argument is in some sense akin to what you depicted (but uses the projective tensor product of Fréchet spaces instead of the algebraic one — surely the algebraic one is not the "correct" one in this case).

2. For every $r\in\mathbb N$, the sheaf $\Omega_{(-)}^r$ has vanishing higher cohomology, since it is a $C^\infty(-)$-module (for this step, you really need the paracompactness).

3. Combining these two points, and apply the de Rham–Weil theorem, we see that the de Rham cohomology, as an object in the derived category $D(M)$ of abelian sheaves, is equivalent to the cohomology $R\Gamma(M;\mathbb R)$ of the constant sheaf $\underline{\mathbb R^{\operatorname{disc}}}$.