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What does it mean to say that a curve in $\mathbb{R}^2$ is embedded? I think a curve in $\mathbb{R}^3$ is embedded if it lies on a plane, but what does it mean in 2d? I searched everywhere but I can't find an answer.

Also, is there a simple way of seeing an immersive curve in $\mathbb{R}^2$?


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The difference between an immersion and an embedding is that the former does not need to be injective. See – Willie Wong Jun 26 '12 at 11:31
Also, even when an immersion is injective, it need not be a homeomorphism onto its image. – Zhen Lin Jun 26 '12 at 11:45
And a curve embedded into $\mathbb{R}^3$ need not lie on a plane - for example, a helix. – Neal Jun 26 '12 at 12:26

In the smooth context, an embedding is a diffeomorphism onto its image. A curve in $\mathbb R^2$ is really a smooth map $\gamma:\mathbb R\to \mathbb R^2$. This map must have a smooth inverse $\gamma^{-1}: \gamma(\mathbb R)\to \mathbb R$ in order for the curve to be embedded. In particular, this requires $\gamma'$ to be nonzero (otherwise the inverse can't be smooth). An embedded curve can look like this:


Having an immersed curve asks only for nonzero derivative. Being a diffeomorphism is not required. An immersed curve can look like this:


To make the distinction trickier, an injective immersion can fail to be an embedding. (As Zhen Lin said.) The figure below shows an immersed line: the immersion is such that the limits $\lim_{t\to \pm\infty}\gamma(t)$ are the "intersectinn" point. There is no actual intersection: the curve passes through the center of the figure only once. This is an injective immersion. Not an embedding, because the inverse map $\gamma^{-1}$ is not even continuous.

injective immersion

I think a curve in $\mathbb R^3$ is embedded if it lies on a plane

This is totally wrong (as Neal pointed out). Being embedded into any space means the same thing: diffeomorphism onto the image. Or homeomorphism, when we are in the topological setting.

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