**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.



Non-uniqueness of loop-erasure for continuous-time curves
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**](http://dx.doi.org/10.1007/s00220-011-1234-9), Commun. Math. Phys. 303, No. 3, 709-720 (2011).


