Suppose I have a smooth curve [itex]\gamma:[0,1] \to M[/itex], where [itex]M[/itex] is a smooth [itex]m[/itex]-dimensional manifold such that [itex]\gamma(0) = \gamma(1)[/itex], and [itex]\hat{\gamma}:=\gamma|_{[0,1)}[/itex] is an injection. Suppose further that [itex]\gamma[/itex] is an immersion; i.e., the pushforward [itex]\gamma_*[/itex] is injective at every [itex]t\in [0,1][/itex].(adsbygoogle = window.adsbygoogle || []).push({});

Claim: The image set [itex]\gamma([0,1])[/itex] is a smooth embedding of the unit circle [itex]S^1[/itex].

This seems intuitively obvious to me -- I can "bend and stretch" a circle into any "nice" closed curve (and vice versa). However, I'm having trouble proving this, even for the special case where [itex]M=\mathbb{R}^n[/itex].

Here is how I've been approaching this: If [itex]\beta:[0,1]\to \mathbb{R}^2[/itex] is a smooth parametrization of the unit circle such that [itex]\hat{\beta}:=\beta|_{[0,1)}[/itex] is an injection, [itex]\beta(0) = \beta(1)[/itex], [itex]\dot{\beta}(t) \not = 0[/itex] for all [itex]t[/itex], and [itex]\beta([0,1]) = \hat{\beta}([0,1)) = S^1[/itex], then [tex]\hat{\beta}^{-1}:S^1 \to [0,1)[/tex] and [tex]\alpha \circ \hat{\beta}^{-1}:S^1 \to C [/tex] where [itex]C:= \gamma([0,1])[/itex]. It seems that if I can show that [itex]\alpha \circ \hat{\beta}^{-1}[/itex] is smooth with an everywhere-injective pushforward, I'd be done. However, I can't figure out how to do this as [itex]\hat{\beta}^{-1}[/itex] isn't even continuous, so I don't think it is differentiable.

Any ideas? I feel like I'm missing something obvious. What little differential geometry (and topology) I know is self-taught, so maybe there is a standard trick I haven't learned.

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# Is every smooth simple closed curve a smooth embedding of the circle?

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