The question that follows is related to the possible values of the index of a simple closed curve. Actually, the question that follows is equivalent to the fact that the index of a simple closed curve can only be equal to -1,0 or 1. For, if the index was n>1, then there would exist a simple closed curve $\gamma(t)=r(t)e^{i\theta(t)}$, $t\in [0,1]$, with $r(0)=r(1)$ and $\theta(1)-\theta(0)=2n\pi$. hence the function $$f(t)=((r(t),s(t)),\quad \text{where}\quad s(t)=\frac{\theta(t)}{2\pi}, $$ would satisfy the assumptions of the proposed question, and if the answer to the question is positive, the curve is not going to be simple.
Question. Let $f : [0,1] \to \mathbb R^2,\,$ be a continuous map such that $$ f(0)=(0,0) \quad\text{and}\quad f(1)=(0,n), $$ where $n\in\mathbb N$, $n>1$. Are there always $x_1,x_2\in [0,1],\,$ such that $$ f(x_2)-f(x_1)=(0,1)? $$
There exists a positive answer for $n=2$. See A question about continuous curves in $\mathbb{R}^2$. However, the proof for $n=2$ doesn't seem to be easily generalized for $n>2$.
Also, the answer would be negative if $n$ was not an integer. (See Average pace and horizontal chords)
Hints and/or references are welcome.
