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?


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    $\begingroup$ Some topological manifolds have no smooth atlases at all. See math.stackexchange.com/questions/408221/… for instance. $\endgroup$ Jan 29, 2020 at 2:38
  • $\begingroup$ 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. $\endgroup$ Jan 29, 2020 at 3:19

1 Answer 1


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.

(Probably there exists a more elementary counterexample than this, but I would not be surprised if any example is fairly hard and in particular well beyond what you could hope to understand if you are just learning about smooth manifolds for the first time.)


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