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I always had this doubt about the topologies one can introduce into a smooth manifold:

Let's say I have a smooth manifold $M$. We can equip this manifold with a topology induced by the smooth structure as follows: we say that a subset $O\subset M$ is open if and only if $\varphi(O\cap U)$ is an open subset of the euclidean space for each chart $(U,\varphi)$ on the smooth structure. Let's call this topology $\tau_M$.

Conversely, it is usual to start our standard examples of smooth manifolds as topological manifolds (that is, topological spaces which are locally euclidean and Hausdorff, second-countable depending on what kind of property one desires) and equip this manifold with a smooth structure where each admissible chart domain is an open subset of the above topology and the chart maps are homeomorphisms. Let's call this topology $\tau$.

Suppose $M$ is just a topological manifold equipped with a smooth atlas such that each chart domain is an open subset and the chart maps are homeomorphisms as above. In this situation, we can show that the topologies $\tau_M$ and $\tau$ coincide.

For instance, if $O\in\tau$, then $\varphi(O\cap U)$ is open for each admissible chart $(U,\varphi)$ since it is the intersection of both open subsets $O$ and $U$ with respect to $\tau$, hence $O\in\tau_M$. Conversely, if $O\in\tau_M$, then for a given admissible chart $(U,\varphi)$ we have that $\varphi(O\cap U)$ is open, and in particular, $(O,\varphi\lvert_{O\cap U})$ is an admissible chart, and therefore $O=\varphi^{-1}(\varphi(O))$ is open (with respect to $\tau$), since $\varphi$ is an homeomorphism.

However, it is known that we not always can transform a topological manifold into a smooth manifold. So, what am I missing here? What's the obstruction to start with a topological manifold structure and define smooth charts with open domains on this given topology? What is the flaw in the argument above?

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    $\begingroup$ The flaw is that you did not present any argument at all. You did not explain how to find a smooth atlas on a topological manifold. $\endgroup$ Feb 24, 2023 at 21:57
  • $\begingroup$ I'm supposing we can find a smooth atlas with the given properties, which is the case for the textbook examples of smooth manifolds: spheres, projective spaces, etc. In this case, the topological manifold topology and the topology induced by the atlas should coincide, right? I've made an edit to show what I'm asking more clear. $\endgroup$
    – kindaichi
    Feb 24, 2023 at 22:02
  • $\begingroup$ There is no argument from you why a topological manifold admits a smooth structure $\endgroup$ Feb 24, 2023 at 22:20
  • $\begingroup$ I'm not claiming that. I'm supposing I have a topological manifold equipped with a smooth structure and I want to know if the topology induced by the atlas coincide with the first topology. $\endgroup$
    – kindaichi
    Feb 24, 2023 at 22:28
  • $\begingroup$ It seems that the only real question in your post is "What's the obstruction to start with a topological manifold structure and define smooth charts with open domains on this given topology?" Answers to this question are known but involve math far beyond what you have learned. An interesting challenge for you is to contemplate the case of 1-dimensional topological manifolds. Why do topological 1-dimensional manifolds admit a (unique up to an isomorphism) smooth structure? $\endgroup$ Feb 25, 2023 at 0:37

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Your reasoning seems to be of the form:

Suppose that $(M,\tau)$ is a topological manifold and $A$ is a smooth atlas on $M$ which is ``compatible'' with the topology $\tau$ on $M$. Then $\tau$ is equal to the topology on $M$ induced by $A$.

But if someone hands you a topological manifold $(M,\tau)$ and nothing else, this statement tells you nothing about whether a smooth atlas that is ``compatible'' with $\tau$ exists.

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  • $\begingroup$ Yes. Thank you. I was thinking about the situations where it is clear that there is a smooth structure since you explicitly present one (e.g., $S^2$). But it is true that in the pathological case there might simply not be a smooth structure at all. I think this clarified it for me. $\endgroup$
    – kindaichi
    Feb 24, 2023 at 22:47
  • $\begingroup$ @kindaichi I wouldn't mind the points from an accepted answer if you found it helpful ;) $\endgroup$
    – user934527
    Feb 27, 2023 at 16:18

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