Contravariant functor $\Omega^*$ from category of smooth manifolds to commutative differential graded algebras

Question

I thought I understood the Mayer-Vietoris Sequence in the context of De Rham cohomology, but I have realized that I have taken something for granted in the set up of the proof, and I can't quite figure it out.

Let $M=U\cup V$ be a smooth manifold, with $U$ and $V$ open, we have the following sequence of maps:

$$U\cap V \longrightarrow U\coprod V\longrightarrow M$$

Then the $\Omega^*$ functor (taking each smooth manifold to it's commutative differential graded algebra of forms) gives us the following sequence:

$$\Omega^*(M)\longrightarrow \Omega^*\left(U\coprod V\right)\longrightarrow \Omega^*(U\cap V)$$

where the maps are given by the pullbacks of the restrictions in the previous sequence. Now, what I do not understand was replacing the middle term with:

$$\Omega^*(U)\oplus\Omega^*(V)$$

as I am not sure how to guarantee that $\Omega^*(U\coprod V)$ is the product/coproduct in the category of commutative graded differential algebras over $\mathbb{R}$.

In particular, the product and the coproduct are the same in this category, so I naively thought that using the universal property of coproducts, and then applying the $\Omega^*$ would yield an object which satisfies the universal property of the product of $\Omega^*(V)$ and $\Omega^*(U)$. However, this argument only works if every commutative differential graded algebra can be written as the De Rham complex of any smooth manifold, which I find very hard to believe.

Is there another way to see the isomorphism $\Omega^*(U)\oplus \Omega^*(V)\cong \Omega^*(U\coprod V)$?

Answer 1

First, it is an important step to prove that the coproduct in the category of dg algebras is as we might expect (the standard direct sum and the direct sum of the differential).

There are natural restriction maps $\Omega^*(U\sqcup V)\to \Omega^*(U), \Omega^*(V)$ given by restricting the domain of the relevant section so we yield a map $\Omega^*(U\sqcup V)\to \Omega^*(U)\oplus \Omega^*(V)$. You can either provide an inverse to this map (try gluing sections over $U$ and $V$ together) or show indirectly that it yields an isomorphism.