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Question. Is there a continuous curve in the plane that has a non-unique loop-erasure?

Here is the definition of a loop-erasure. A continuous curve $Y:[c,d]\to\mathbb R^2$ is a loop-erasure of a curve $X:[a,b]\to\mathbb R^2$ if there exists an increasing and right-continuous function $w:[c,d]\to [a,b]$ such that:

  • $w(c)=a$,
  • $X(w(d))=X(b)$,
  • $Y(t)=X(w(t)), \forall t$,
  • For every $T$, the image of the curve $Y(t), c\leq t\leq T$ does not intersect the image of the curve $X(s), w(T)<s\leq b$.

Note that if $w$ has a jump at time $t$, then one should have $X(w(t^-))=X(w(t))$. This corresponds to erasing a loop in $X$.

Update. Two loop-erasures are equivalent if they have the same image.

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  • $\begingroup$ Couldn't two erasers of $X$ erase different loops of $X$? $\endgroup$ – Jochen Wengenroth Sep 24 at 12:02
  • $\begingroup$ @JochenWengenroth No, the definition implies that $Y$ should not intersect itself. So, for instance, if $X$ has finitely many self-intersections, then one should erase all of the loops and there is a unique loop-erasure. $\endgroup$ – Ali Khezeli Sep 24 at 15:40
  • $\begingroup$ Okay, I see now. $\endgroup$ – Jochen Wengenroth Sep 24 at 15:52
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The paper that Iosif Pinelis mentioned in his answer has an example to this problem: Consider the compact space obtained by adding $\pm\infty$ to the strip $\{ z\in \mathbb C: 0\leq \mathrm{Im}(z)\leq 1\}$. Consider the curve that connects the integer points of this set (by segments) in the following order: $\ldots, n, n+1, n+i, n+i+1, n+1, n+2,n+i+1,n+i+2,n+2,\ldots$. Then there are two non-equivalent loop-erasures of this curve: One has image $\mathbb R\cup\{\pm\infty\}$ and one has image $(\mathbb R+i)\cup\{\pm\infty\}$.

Zhan, Dapeng, Loop-erasure of planar Brownian motion, Commun. Math. Phys. 303, No. 3, 709-720 (2011).

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    $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ – András Bátkai Sep 24 at 7:46
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    $\begingroup$ @AndrásBátkai According to your comment, I edited my answer and included the example here. $\endgroup$ – Ali Khezeli Sep 24 at 15:39
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$\newcommand\tt{\tilde t}\newcommand\tY{\tilde Y}\newcommand\tw{\tilde w}$A loop-erasure is never unique whenever it exists (see e.g. this paper for a nontrivial example of such an existence):

Zhan, Dapeng, Loop-erasure of planar Brownian motion, Commun. Math. Phys. 303, No. 3, 709-720 (2011). ZBL1217.60076.

Indeed, if a continuous curve $Y\colon[c,d]\to\mathbb R^2$ is a loop-erasure of a curve $X\colon[a,b]\to\mathbb R^2$, witnessed by a function $w\colon[c,d]\to[a,b]$, then for any real $s$ the curve $[c-s,d-s]\ni\tt\mapsto\tY(\tt):=Y(s+\tt)$ is also a loop-erasure of the curve $X$, witnessed by the function $[c-s,d-s]\ni\tt\mapsto\tw(\tt):=w(s+\tt).$

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  • $\begingroup$ I forgot to say that such examples are considered equivalent. I will update the question. But the paper that you mentioned has already an example for my question! Thanks. $\endgroup$ – Ali Khezeli Sep 24 at 5:12

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