# Is there a converse to “knowing a good cover” → “knowing cohomology”?

When looking at smooth manifolds, knowing a “nice” cover of our space enables us to calculate the De Rham cohomology via e.g. the Meyer-Vietoris sequence. For instance, $$M := \{(x,y)\mid x^2+y^2\in (1,2) \} \hookrightarrow \mathbb R^2$$ can be covered by two sets $$A\simeq B\simeq \mathbb R^2$$ whose intersection is $$\mathbb R^2 \sqcup\mathbb R^2$$. Because we know the cohomology of $$\mathbb R^2$$, we can utilize this cover to calculate the cohomology of $$M$$.

Obviously, there are multiple ways to cover our set $$M$$ (e.g. by just taking more and more sets with equally trivial intersection), but this one is particularly small (number of sets) and only has sets diffeomorphic to $$\mathbb R^n$$ and simple intersection, i.e. intersection diffeomorphic to disjoint unions of $$\mathbb R^n$$.

However, if we knew the cohomology of $$M$$, are there any ways to assert existence of such a „nice“ cover?

I know that the cohomlogy is pretty „weak“ in the sense that it does not „see“ changing to a non-diffeomorphic manifold if it has the same homotopy type, but I'm not even sure if having the same cohomology implies having the same homotopy type. So it might be necessary to demand knowledge of the original manifold like dimension or what the minimal $$n$$ is such that it can be embedded into $$\mathbb R^n$$.