I have perused the related questions on this site, and was unable to find a formal proof of the fact stated in the title. Essentially, I have two questions:
- Is it a fact that if $M$ is a submanifold of $N$, there must exist an embedding of $M$ into $N$?
- If the above is true, how would one show that there is no embedding of $V := \{(x,|x|)\ |\ x\in \mathbb{R}\}$ into $\mathbb{R}^2$?
In my attempts, I have been assuming (1) and beginning along the lines of "suppose $u:V\rightarrow \mathbb{R}^2$ is an embedding. Then since $V$ is a one-dimensional manifold (it can be covered by the single chart $(V,(x, |x|) \mapsto x)$), it is homemorphic to $\mathbb{R}$; in particular, we can assume WLOG that a neighbourhood of $(0,0)$ is homeomorhpic to a neighbourhood of the origin of $\mathbb{R}$. We can therefore assume that (in a nbhd. of $(0,0)$), $u:(-\epsilon,\epsilon)\rightarrow\mathbb{R}^2$. Then $u(x)=(x,|x|)$ necessarily..." If my proof is right upto here, it is straightforward to then show that the derivative of $u$ vanishes and so it cannot be an embedding. I'm not sure of its correctness though, so I thought I would post here for feedback.
Thanks!
As Mike Miller appointed, you need to show that there is no $u:(-\epsilon,\epsilon)\to\mathbb{R}^2$ smooth immersion with $u((-\epsilon,\epsilon))=V$.
To see this, just write $u(t)=(u_1(t),|u_1(t)|)$ and suppose $u(0)=(0,0)$.
The derivative of $|u_1|$ at $t=0$ (which exists because $u$ is smooth) is:
$$\lim_{h\to 0}\dfrac{|u_1(h)|}{h}.$$
If $h>0$ that number is $|u_1'(0)|$, and if $h<0$ that number is $-|u_1'(0)|$. Hence $u_1'(0)=0$ and $u'(0)=(0,0)$. Thus $du_{0}=0$ and $u$ can't be immersion.