Let $M$ be a smooth manifold. Let $C \subseteq M$ be a simply connected closed set. Is it always possible to find a simply connected open set $U \subseteq M$ such that $C \subseteq U$?
2026-03-27 14:03:32.1774620212
Simply connected neighbourhood of a simply connected closed set
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No. Consider $M=\mathbb{R}^2\backslash\{0\}$ and let $C$ be the topologist's sine circle:
around the origin $0$. It is simply-connected. Since you cannot cross from the right side to the left side via upper half. And so the image of any path is contained in an arc.
The point is that any open neighbourhood of that space contains $S^1$ (up to deformation anyway) which cannot be contracted due to $0$ not belonging to $M$.