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Define $\phi(t)=(e^{it},e^{i\sqrt{2}t})$ which goes from the real line to the Torus. I understand that the differential of this map is injective and so it is an immersion, but what I don't understand is why $\phi(\mathbb{R})$ is not an embedding? Any hints/remarks will be much appreciated.

Notes: I read that to establish an embedding one needs to show that $\mathbb{R}$ is diffeomorphic to its image $\phi(\mathbb{R}).$ So I am guessing one needs to show that such a diffeomorphism does not exist. But this does not seem straightforward to me.

Student
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  • Can you state the definition of an embedding? How is it different from that of an immersion, and so where would things go wrong? – While I Am Jan 22 '22 at 17:25
  • I am just using the definition of embedding from Wikipedia A smooth embedding is an injective immersion $f : M → N$ that is also a topological embedding, so that $M$ is diffeomorphic to its image in $N.$ – Student Jan 22 '22 at 17:29
  • However, this is an example in our lectures on lie algebra where the professor did not give a definition so I am not sure what they had in mind. The idea there was to to construct a lie algebra whose corresponding Lie subgroup is an immersion. – Student Jan 22 '22 at 17:29
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    @Zest It is injective. The problem is that its image is dense in the torus so that it is not an homeomophism onto its image – Didier Jan 22 '22 at 17:39
  • Ah, yes, i wasn't careful enough. My apologies. – Zest Jan 22 '22 at 17:41
  • @Didier Could you please explain why is the image dense and how does that imply it is not a homomorphism onto its image? – Student Jan 22 '22 at 17:44
  • Or this one – Arctic Char Jan 22 '22 at 18:09

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