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Let $\alpha:[0,1/2] \rightarrow \mathbb{R}^n$ be a continuous, injective map for some $n \geq 2$. Can we extend $\alpha$ to a continuous map $\beta:[0,1] \rightarrow \mathbb{R}^n$ sastisfying $\beta(0)=\beta(1)$ and such that $\beta(t_1) \neq \beta(t_2)$ for every two distinct points $t_1, t_2$, with at least one of them belonging to $(0,1)$?

In other terms, given a simple arc, can we join its end-points so to form a simple closed curve? The answer should be positive, but I have no idea of a possible general proof of this fact, so any help is welcome. Thank you very much in advance for your kind attention.

Maurizio Barbato
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    What you tried proving it? What background do you have? Any more context on this question? – Arctic Char Jan 01 '25 at 17:15
  • The question came up out of my curiosity, while I was solving a problem. My background in general topology is at the level of an undergraduate course. – Maurizio Barbato Jan 01 '25 at 17:20
  • I think for $n = 2$ this it true, but by far non-trivial, see here for a discussion. I guess for higher dimension it is false, see here. – Ulli Jan 01 '25 at 17:35

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The linked question/answer does not contain an actual answer (in the case $n=2$), only a suggested strategy without an actual proof. There are several ways to argue. One is by reducing the problem to the Schoenflies theorem: Let $A\subset S^2$ be a (simple) topological arc (aka Jordan arc) with the end-points $p, q$. Take the 2-fold branched cover $\pi: S^2\to S^2$ ramified over the points $p, q$. Then $A$ lifts homeomorphically to two disjoint (simple) arcs $A_1, A_2\subset S^2$ with the end-points at $\pi^{-1}(\{p,q\})=\{x,y\}$. Thus, $C=A_1\cup A_2\subset S^2$ is homeomorphic to $S^1$, hence, by Schoenflies, there exists a homeomorphism $h: S^2\to S^2$ sending $C$ to a round circle $C'$. Take a chord $c'$ in $S^2$ connecting $h(x), h(y)$ and otherwise disjoint from $C'$. Then $h^{-1}(c')$ is a simple arc connecting $x, y$ and otherwise disjoint from $C$. Now, take $\pi(c)=b$: It is a simple arc in $S^2$ connecting $p, q$ and otherwise disjoint from $A$. This is the arc you asked for. There are other arguments as well, for instance, using the Riemann mapping theorem: mapping the open unit disk $D$ via a conformal map $f: D\to S^2\setminus A$, taking a chord $c$ in $D$, mapping $c$ via $f$, and then using the Caratheodory-Torhorst extension theorem (without the need to use a branched cover) to argue that the open arc $f(c)$ extends continuously to the end-points.

Now, in higher dimensions, this is still true. The proof is a bit long and uses the fact that topological arcs in $\mathbb R^n$, $n\ge 3$ do not locally separate. See my Mathoverflow answer here. It is written for totally disconnected subsets but the proof is the same for (simple) arcs.

Moishe Kohan
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  • Dear @MoisheKohan I am quite astounded by your answer! I suspected this question was not trivial, but I could not imagine that an answer should have required such advanced notions as branched coverings, conformal mappings or a remarkable result like the Riemann mapping theorem! This is one of the aspects of mathematics that amaze me more: one cannot ever know in advance where simple questions will lead! It is really a wonderland! – Maurizio Barbato Jan 02 '25 at 12:45
  • Thank you very very very ... much for having devoted some of your precious time to answer this question in detail! May HaShem be blessed, for He has given us the joy of understanding something of His mysterious creation and that of sharing this comprehension with others! And Chanukkah sameach for this last festival day! – Maurizio Barbato Jan 02 '25 at 12:47