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Show that the curve $t\rightarrow (t,t^2,t^3)$ embeds $\mathbb{R}$ into $\mathbb{R^3}$. Find two independent functions that globally define the image. Are your functions independent on all of $\mathbb{R^3}$ or just an open neighborhood of the image?

My Try:

To show that it is an embedding I must show it is continuous and a homeomorphism onto its image. I think it is obvious. But I do not have any clue on how to approach the rest. Can anybody please help me?

Extremal
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1 Answers1

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"I think it is obvious": the continuous part maybe, but the fact that it is a homeomorphism with its image is arguably not self-evident. Luckily, the inverse in this case is rather simple (what is it?), so it is easy to show that it is a homeomorphism with its image.

Probably the "embedding" part also wants you to show that it is an immersion since you tag in differential topology, so you should also check that.

Now, about finding two independent functions: you want $x^2=y$ and $x^3=z$ in that curve, correct? Use that to define the functions.

Aloizio Macedo
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  • Thanks. The inverse is the projection map of the first coordinate. That’s why I said it is obvious. – Extremal Feb 07 '18 at 17:38
  • Regarding independent functions, as I have to find two, each should be a function from $\mathbb{R}$ to $\mathbb{R^2}$. But the image is in $\mathbb{R}^3$. So I’m still confused. – Extremal Feb 07 '18 at 17:49
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    @Convex I believe you are misunderstanding the problem. I think you are supposed to find two functions $f,g: \mathbb{R}^3 \to \mathbb{R}$ such that the image of the curve $\gamma$ is $f^{-1}({c}) \cap g^{-1}({c})$. – Aloizio Macedo Feb 07 '18 at 17:58
  • So far what I could guess is $f(x,y,z)=|y|^{1/2}$ and $g(x,y,z)=z^{1/3}$. But then $f^{-1}({c})\cap g^{-1}({c})={(x,c^2,c^3)| x,c\in \mathbb{R}}$ which is not exactly the curve. It seems that I must involve $x$ somehow in $f,g$. But how could that be possible? – Extremal Feb 08 '18 at 05:01
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    @Convex When I write "$c$", I mean a constant. For example, the functions $f,g: \mathbb{R}^3 \to \mathbb{R}$ given by $f(x,y,z)=x^2+y^2+z^2-1$ and $g(x,y,z)=z$ determine the unit circle on the $x,y$ plane when you consider $f^{-1}({0}) \cap g^{-1}({0})$. You are supposed to do something similar for this case. – Aloizio Macedo Feb 08 '18 at 05:44
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    I see. I have misunderstood again :) Thank you for the clarification. – Extremal Feb 08 '18 at 06:28