2

Note: This is a followup to another question of mine. The statement is almost the same; the only difference is that I ask for injective path-connectedness instead of arc connectedness.

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 injectively immersively path-connected, i.e. that we can choose $\gamma$ to be injective?

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, and we get homeomorphic, not just injective paths). As shown in the answers to the previous question, in general, arc connectedness does not follow.

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

Eric Wofsey
  • 330,363
tomasz
  • 35,474

1 Answers1

5

Take the union of the segment $[0,1]\times\{0\}\subset\mathbb{R}^2$ and the circle centered at $(2,0)$ with radius $1$ and add an extra copy of the point $(0,0)$ (as in the line with two origins) to get a space $X$. Then $X$ is immersively path-connected: to get an immersed path between the two origins, go out along the segment starting from one of them, then around the circle, and then back along the segment to the other one. But there is no injective path between the two origins.

Eric Wofsey
  • 330,363
  • Thanks, this works. I was actually thinking of something very similar before, only also doubling $(1,0)$ and changing the topology so that the two "branches" of the circle connect to separate copies of $(1,0)$. I don't know why I thought it does not work, but now it seems to me that that is fine as well. It has the additional advantage of being a manifold. – tomasz Nov 25 '23 at 01:34