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Edit: I've asked a followup question.

Suppose $X$ is a topological space such that any two points $x_0,x_1\in X$ are connected by an immersive path, i.e. there is a locally homeomorphic embedding $\gamma\colon [0,1]\to X$ such that $\gamma(0)=x_0$ and $\gamma(1)=x_1$.

Does it follow that $X$ is arc-connected, i.e. that we can choose $\gamma$ to be an actual homeomorphic embedding?

Note that we can assume without loss of generality that $X$ is the image if an immersive path, in particular that it is a compact space which is covered by (finitely many, by compactness) arcs. Further, the conclusion is true if $X$ is Hausdorff (in fact, just being path-connected is sufficient in this case).

(This is sort of motivated by my ruminations related to this question.)

Eric Wofsey
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tomasz
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  • Can you give a pointer to a proof that this is true if $X$ is Hausdorff, please. – Rob Arthan Nov 24 '23 at 21:07
  • Well, a Hausdorff path-connected space is arc-connected. This is well-known. See this post for a careful proof. – tomasz Nov 24 '23 at 22:13
  • @RobArthan: Ah, perhaps you were confused because I wrote "local homeomorphism" and "homeomorphism" where I meant a an embedding. This is fixed now. – tomasz Nov 24 '23 at 22:16
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    Actually, I didn't know that "well-known" proof, so many thanks for the pointer. I'll be interested to see your comments on my tentative answer to your question. – Rob Arthan Nov 24 '23 at 22:34
  • @RobArthan: I apologise, it just occured to me that I might have come across as a bit condescending. It was not my intent, it's just that I've heard about this quite a few times in different places (although to be honest, I've never actually read or written a full proof of this fact). – tomasz Nov 24 '23 at 23:32
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    No need to apologise: we all learn that "well-known" and "easy to see" and such like phrases in mathematical discourse often have to be taken with a pinch of salt. – Rob Arthan Nov 24 '23 at 23:34

2 Answers2

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Let $X=S^1\cup \{1'\}$ (i.e. the circle with a double of $1$). Then any pair of distinct points other than $1$ and $1'$ can be connected by an arc, and $1$ and $1'$ can be connected by an immersive path, namely we take a path from $1$ all the way around the circle, only coming back to $1'$ instead of $1$. The resulting path is clearly locally a homeomorphic embedding, but it is not a homeomorphic embedding because its image is not Hausdorff.

In fact, this $X$ is even a compact path-connected smooth manifold. It is also injectively path-connected.

tomasz
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3

This is a suggestion at an answer that is a bit too long to write up as a comment. Apologies if it is baloney. Let $A = [0, 1] \times \{0, 1\} \subseteq \Bbb{R}^2$ and take all sets of the form $(a, b) \times \{i\} \cup \{x\} \times \{ 1 - i\}$, where $0 \le a < x < b \le 1$ and $i \in \{0, 1\}$ as the subbasis of a topology on $A$. This is along the lines of the line with two origins, but now I am arranging for every point in the interior of the unit interval to have a doppelganger. Now let $B$ be $\{1\} \times [0, 1]$ equipped with the usual topology and take $X$ to be $A \cup B$ with the topology induced by the topologies given for $A$ and $B$. Then any two points in $X$ are connected by an immersive path (you can just ignore the doppelgangers in a neighbourhood of any point), but no point with a doppelganger is connected by an arc to its doppelganger (as the image of an arc must be Hausdorff, but no subset of $X$ containing a point and its doppelganger is Hausdorff).

Rob Arthan
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  • How do you get an immersive path from a point to its doppleganger? – tomasz Nov 24 '23 at 23:40
  • By ignoring the doppelgangers in neighbourhoods of point with doppelgangers. I.e., just take the same path that you would take if we'd given $A$ the usual topology. – Rob Arthan Nov 24 '23 at 23:55
  • Ah, I see, this is where $B$ comes in. – tomasz Nov 25 '23 at 01:04
  • Yes, this works. In fact, I was thinking of a similar and simpler example, namely a circle with a single doubled point, although for some reason I thought it does not work. I will post it as a separate answer. – tomasz Nov 25 '23 at 01:08
  • Ah, I see now why I thought that. The question I was ruminating on asked about injective paths, not arcs. Both the spaces you and I consider are in fact injectively path-connected. It seems that there is one more question I'll have to ask. :-) – tomasz Nov 25 '23 at 01:14
  • Here's the followup question, in case you're interested: https://math.stackexchange.com/q/4813832/30222 – tomasz Nov 25 '23 at 01:22