Let's say I have a smooth manifold $M$ and some subset $U$ of $M$, equipped with the subspace topology. The inclusion map $i:U\to M$ is a topological embedding. Is there always a smooth structure on $U$? Is it unique?
Topological embedding and smooth structure
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general-topology
differential-geometry
differential-topology
smooth-manifolds
1 Answers
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No, there typically is not any smooth structure on $U$: $U$ will usually not even be locally homeomorphic to $\mathbb{R}^n$ for any $n$. For instance, consider $M=\mathbb{R}$ and $U=\mathbb{Q}$. Even if $U$ is a topological manifold, there is no reason that it must "inherit" any smooth structure from $M$. For instance, every (second-countable) topological manifold embeds topologically in $M=\mathbb{R}^N$ for some $N$, but some topological manifolds admit no smooth structures.
If $U$ is open in $M$, then it does have a canonical smooth structure, obtained by just restricting all the charts of $M$ to $U$. (You need $U$ to be open for this to work, so that the images of the charts will still be open subsets of $\mathbb{R}^n$.) You can show that this is the unique smooth structure on $U$ for which the inclusion map is a smooth embedding.