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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.

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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.

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    $\begingroup$ I'm not sure why the last sentence is right "Then the result is obvious". The problem is that your coordinate chart doesn't preserve circles. $\endgroup$ – Anthony Quas Feb 11 at 22:35
  • $\begingroup$ I think you are assuming that $\gamma$ is a homeomorphism, but the OP doesn't say this. Clearly the claim is false under the assumptions the OP stated. (I'm also wondering why $\dim \mathcal M =2$, though perhaps the use of the word circle suggests this.) $\endgroup$ – Christian Remling Feb 11 at 22:36
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    $\begingroup$ @AnthonyQuas $\mathcal M$ is just a manifold, so doesn't have an intrinsic notion of a circle (like Riemannian manifolds, say). I suspect OP means topological circles. $\endgroup$ – Wojowu Feb 11 at 22:38
  • $\begingroup$ @Wowoju: I don't think so: notice that the OP talks about the radius of the circle... $\endgroup$ – Anthony Quas Feb 11 at 22:39
  • $\begingroup$ @AnthonyQuas My apologies for not being precise with the terminology. The precise version of the question is given in the answer above by Piotr Hajlasz (though it should also add "there exists an r"). I meant radius of the circle in the coordinate system containing $\gamma(0)$. So yes, upstairs in the manifold it is indeed a topological circle. $\endgroup$ – Rahul Sarkar Feb 11 at 23:18

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