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It seems to not be generally true that a subset $U$ of a simply connected space $E$, is simply connected. The most obvious example perhaps is that $\mathbb{R}^2$ is simply connected, but that $\mathbb{R}^2 \setminus\{(0,0)\}$ is not simply connected (take any loop going around $(0,0)$).

Are there further assumptions one can impose on either $E$ (e.g. locally path-connected/connected) and/or $U$ under (e.g. component/path-component of $E$) under which one can say that $U$ will be simply connected?

Moishe Kohan
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Ben123
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    As you note, you can poke a hole in pretty much any simply connected set and arrive at something that is not simply connected. But poking a hole in a set doesn't have that much effect on other topological properties, particularly local properties (which generally won't even see the hole). I don't know how you can do much better than "A subspace of a simply connected space is simply connected if and only if it is simply connected." – Xander Henderson Nov 23 '24 at 17:32
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    If you look in $\mathbb R^3$, and you set the vast complexity of non-simply connected subsets next to the vast complexity of simply connected subsets, perhaps an answer of No becomes appropriate to your question of "Are there further assumptions..." I would say that this question is really too vague to have any more reasonable answer. – Lee Mosher Nov 23 '24 at 17:36
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    See for instance my answer here. But try to narrow down your question to something answerable. – Moishe Kohan Nov 23 '24 at 17:39
  • See also the answer here. But keep in mind that for every $n$ there is a Cantor subset $K\subset \mathbb R^n$ whose complement is not simply connected. – Moishe Kohan Nov 23 '24 at 17:58
  • Note that a necessary condition is that the space is hereditarily connected. Would it be sufficient? – Steven Clontz Nov 24 '24 at 16:29
  • Hereditarily connected spaces are path connected. Are they simply connected to? https://topology.pi-base.org/spaces?q=Hereditarily+connected%2B%7ESimply+connected – Steven Clontz Nov 24 '24 at 16:31
  • @StevenClontz: Your edits have changed the question to a completely different question. – Lee Mosher Nov 24 '24 at 19:47
  • I don't think so. The spirit of the question was about guaranteeing when subspaces of simply connected are simply connected. – Steven Clontz Nov 24 '24 at 20:26
  • Of course, OP can correct me if I'm mistaken. – Steven Clontz Nov 24 '24 at 20:29

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I believe you're discussing the hereditarily connected spaces, which are characterized by having a topology totally ordered by the subset relation. https://topology.pi-base.org/properties/P000196 Every subspace of such a space is in turn hereditarily connected and path connected.

I will show a hereditarily connected space has a trivial fundamental group. Given a loop, we may identify a minimum and maximal point with regards to the topology order. We may reparameterize points so the maximum point occurs at ±1 and the minimum point occurs at 0, and each point between is reached at ±t. Finally, we may continuously deform the loop to a point by maxing out the loop at ±(1-t) from 0 to 1.

(On phone so I'm waiving my hands a bit; if I'm not overlooking an error I'll follow up to justify details if no one beats me to out.)

Edit: a more robust answer to this question is here. https://math.stackexchange.com/a/5000543/86887

Steven Clontz
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