# Is every possible chart a member of some maximal smooth atlas?

If $$M$$ is a topological $$n$$-manifold (edit: that admits at least one smooth structure) and I select any open set $$U \subseteq M$$, and I find that there exists some $$\varphi: U \rightarrow \varphi(U)$$ where $$\varphi(U) \subseteq \mathbb{R}^n$$ and $$\varphi$$ is a homeomorphism, then the pair $$(U, \varphi)$$ are a chart on $$M$$. Is it necessarily the case that some smooth structure $$\overline{\mathcal{A}}$$ exists so that $$(U, \varphi) \in \overline{\mathcal{A}}$$?

A little more context. My current understanding is that there are a number of ways to conceptualize a smooth structure: as a maximal smooth atlas, as an equivalence class of smooth atlases, or as a maximal set of mutually compatible charts. I've found a couple of great answers about smooth structures that consider the smooth structure from the perspective of equivalent atlases (Manifold and maximal atlas and Why maximal atlas). And I also understand that given even just one smooth atlas $$\mathcal{A}$$, one can essentially generate a unique maximal smooth atlas $$\overline{\mathcal{A}}$$ such that $$\mathcal{A} \subseteq \overline{\mathcal{A}}$$.

So my question is more from the perspective of the individual charts. I am almost certain that not every open set $$U$$ is suitable to be a domain for a chart. For example, $$M$$ is, itself, an open set. But certainly not every $$M$$ is globally homeomorphic to $$\mathbb{R}^n$$. So, I think it's the case that not every $$U \subseteq M$$ is homeomorphic to $$\mathbb{R}^n$$. But for those $$U$$ that are, is it certainly the case that the chart $$(U, \varphi)$$ is included in some smooth atlas?

Thanks!

• Some topological manifolds have no smooth atlases at all. See math.stackexchange.com/questions/408221/… for instance. Jan 29, 2020 at 2:38
• Ah great point! I'm asking because of a question in Lee's Introduction to Smooth Manifolds: "If $M$ has a smooth structure, show that it has uncountably many distinct ones." So I've edited my post to add the assumption that $M$ does admit a smooth structure. Thanks. Jan 29, 2020 at 3:19

No. Here is one kind of ridiculous way to get a counterexample. By a theorem of Demichelis and Freedman there are uncountably many pairwise non-diffeomorphic small exotic $$\mathbb{R}^4$$s: that is, open subsets of $$\mathbb{R}^4$$ which are homeomorphic to $$\mathbb{R}^4$$ but not diffeomorphic to $$\mathbb{R}^4$$. On the other hand, by a theorem of Cheeger there are only countably many different smooth closed manifolds up to diffeomorphism.
Now take $$M=S^4$$ and let $$U$$ be the complement of a point in $$M$$. There is a homeomorphism from $$U$$ to any small exotic $$\mathbb{R}^4$$ which we can consider as a chart on $$M$$. If all of these charts extended to smooth structures, we would get uncountably many non-diffeomorphic smooth structures on $$S^4$$, since they are not diffeomorphic after removing a point. Since $$S^4$$ is compact this is impossible.