I recently found an interesting question, and I'd like to ask the co-question. I will differ from what is linked by using de Rham cohomology, since its the easiest way for me to articulate my confusion, and it gives a possible relationship to the well-known Mayer-Vietoris sequence.
Suppose we have two open subsets $U$ and $V$ that cover a smooth manifold $M$. We can realise the union $M = U \cup V$ as the pushout of the diagram $V \overset{\iota_2}\leftarrow U \cap V \overset{\iota_1}\rightarrow U$. Here both arrows are given by the inclusion maps. The pushout $U\cup V$ also includes two inclusions $\iota_U: U \rightarrow U \cup V$ and $\iota_V: V \rightarrow U \cup V$ which form a commutative diagram in the obvious way.
Since de Rham cohomology can be seen as a contravariant functor, we can always obtain the diagram $H^{n}(V) \overset{\iota_2^{\ast}}\rightarrow H^{n}(U \cap V) \overset{\iota_1^{\ast}}\leftarrow H^{n}(U)$. Here the maps are given by the pullbacks $\iota_1^{\ast}$ and $\iota_2^{\ast}$ (once descended to cohomology). This diagram, which exists in the category of Abelian groups, has a pullback that is given by the fibred product:
$ H^{n}(U) \times_{H^{n}(U \cap V)} H^{n}(V) := \{ ([\mu],[\eta]) \in H^{n}(U) \times H^{n}(V) \ | \ \iota_1^{\ast}([\mu]) = \iota_2^{\ast}([\eta])\}$.
Similar to the linked question, it seems to be the case that the above pullback and the cohomology $H^n(U \cup V)$ are isomorphic -- I am pretty sure that we can use the universal property of the pullback together with the surjectivity and commutativity of the maps $\iota_1^{\ast}, \iota_2^{\ast}, \iota_U^{\ast}$ and $\iota_V^{\ast}$ to obtain a bijective group homomorphism from $H^n(U \cup V)$ to $H^{n}(U) \times_{H^{n}(U \cap V)} H^{n}(V)$.
With the set-up out of the way, here is my question: If it is the case that $H^n(U \cup V) \cong H^{n}(U) \times_{H^{n}(U \cap V)} H^{n}(V)$, then it ought to be consistent with the computation of $H^n(U \cup V)$ via the Mayer-Vietoris sequence, i.e. there should be some way to read off this result from the Mayer-Vietoris sequence. Is this true?
Here is the answer to my question:
The error in the reasoning comes from the assumption that the homomorphism from $H^n(U \cup V)$ to $H^n(U) \times_{H^n(U\cap V)} H^n(V)$ is an isomorphism. According to the Mayer-Vietoris sequence and the first isomorphism theorem, we will get that: $$\frac{H^n(U \cup V)}{Im(d^*)} \cong \ker{ (\iota^*_{U} - \iota_V^*)} = H^n(U) \times_{H^n(U\cap V)} H^n(V),$$ which does not guarantee that $H^n(U \cup V)$ and $H^n(U) \times_{H^n(U\cap V)} H^n(V)$ agree.