Find a circle that intersects the image of $[0,1]$ in a manifold $\mathcal{M}$ at only 1 point

Question

Let $\gamma : [0,1] \rightarrow \mathcal{M}$ be a continuous map so that $[0,1]$ is homeomorphic to $\gamma([0,1])$, where $\mathcal{M}$ is a manifold (Hausdorff, second countable, and locally Euclidean). Using a chart containing $\gamma(0)$, I think it is always possible to find a circle centered at $\gamma(0)$ that intersects the curve $\gamma([0,1])$ at a single point. Can someone help me prove this?

The intuition I have is that it should be possible to choose a radius $r$ small enough so that $\gamma([0,1])$ intersects the circle exactly once. However I'm not sure how to reason that the curve is not like a "space filling curve" locally around $\gamma(0)$. I know this has to do something with $[0,1]$ (hence $\gamma([0,1])$) being compact, but not sure how the argument should go.

Answer 1

My understanding is that $\operatorname{dim}\mathcal{M}=2$ since we are taking about circles. Also that the precise formulation of the question is:

There is a coordinate system $\phi:U\subset\mathcal{M}\to\mathbb{R}^2$ and $r>0$ such $\gamma(0)\in U$ and the set $$ \phi(\gamma([0,1])\cap U)\cap \underbrace{S^1(\phi(\gamma(0)),r)}_{circle} $$ consists of exactly one point.

Yes, that is possible.

Here is a sketch of the argument. We can assume that we are in a coordinate chart homeomorphic to $\mathbb{R}^2$ (since it suffices to consider a small piece of the curve that is near $\gamma(0)$). Then $\gamma([0,1])$ can be extended to a closed Jordan curve, see https://mathoverflow.net/a/75350/121665. By Schonflies theorem there is a homeomorphism of the chart that maps the Joradn curve to a circle. Therefore you can find a coordinate chart in which your curve near $\gamma(0)$ is an arc of a circle or even a straight segment. Then the result is obvious.