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A polynomial map $f: \mathbb{C} \to \mathbb{C}^2$, $t \mapsto f(t):=(f_1(t),f_2(t))$ is called an embedding of $\mathbb{C}$ in $\mathbb{C}^2$ if $\mathbb{C}$ is isomorphic to its image under $f$, see A. van den Essen, page 2. (By a polynomial map we mean that $f_1(t),f_2(t) \in \mathbb{C}[t]$).

After Example 1 (also in page 2), A van den Essen presents a criterion from differential geometry: $f$ is an embedding if and only if $f'(t) \neq 0$ for all $t \in \mathbb{C}$ and the map $f: \mathbb{C} \to \mathbb{C}^2$ is injective.

Now, replace $\mathbb{C}$ by $\mathbb{R}$ and call a polynomial map as above an embedding of $\mathbb{R}$ in $\mathbb{R}^2$ (= $\mathbb{R}$ is isomorphic to its image under $f$).

Is the differential geometry criterion holds over $\mathbb{R}$?

Any hints are welcome!

Edit: (1) I do not see how this question or this question help in solving my question, although they are somewhat relevant (I am not sure if my question should be tagged also 'real analysis' or 'general topology'). (2) I guess that my question has a trivial positive answer? Namely, such a criterion is also valid over $\mathbb{R}$?

Another Edit: My question has a counterexample $\mathbb{R}[t^2,t+t^3] \subsetneq \mathbb{R}[t]$, but $f(t)= (t^2,t+t^3)$ satisfies the geometric conditions ($f'(t) \neq 0$ and $f$ is injective); please see this question. Therefore, I would like to ask:

Is it possible to find an additional condition (in addition to the two geometric conditions $f'(t) \neq 0$ and $f$ is injective) which will guarantee that $\mathbb{R}[f_1(t),f_2(t)]=\mathbb{R}[t]$? (Perhaps the additional condition will involve the second derivative $f''(t)$?).

Thank you very much!

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  • $\begingroup$ There is an injective immersion ($f'(t)\neq0$) that is not a topological embedding. See the figure in An injective immersion that is not a topological embedding. In my opinion, the author assumes $f$ is a polynomial. Then you can avoid this example. $\endgroup$
    – ChoF
    Mar 19, 2018 at 2:36
  • $\begingroup$ @ChoF, thank you very much for your comment! Unfortunately, I still do not understand what is the answer to my question. If I understand correctly, the example in the link is $f(t)=(\sin{2t},\sin{t})$, which is not a polynomial map (so perhaps my question has a positive answer after all?). $\endgroup$
    – user237522
    Mar 19, 2018 at 3:13
  • $\begingroup$ If $f$ is polynomial, i.e., $f_1(t),f_2(t)\in\mathbb{R}[t]$, then the "if and only if" condition makes sense. Actually ($\Rightarrow$) holds always, but ($\Leftarrow$) may not. In case that $f$ is polynomial, ($\Leftarrow$) also holds, I think. $\endgroup$
    – ChoF
    Mar 19, 2018 at 3:29
  • $\begingroup$ The extra condition you need to get an embedding is properness. Can you prove that a polynomial map $\Bbb R\to\Bbb R^2$ is proper (preimage of compact is compact)? $\endgroup$ Mar 19, 2018 at 5:04
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    $\begingroup$ @user237522 As Ted Shifrin pointed out, it is well-known that "Let $f\colon M\to N$ be a proper injective immersion of differentiable manifolds $M$ and $N$. Then $f$ is an imbedding and $f(M)$ is a regular submanifold and a closed subset of $N$, and conversely." You can find the proof here (p.8). $\endgroup$
    – ChoF
    Mar 19, 2018 at 6:14

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I'm going to write a brief answer here, rather than continuing in comments. The algebraic condition cannot be recovered by a differential topological criterion. Here's the proof.

I will write down a global diffeomorphism (which is definitely not algebraic) of $\Bbb R^2$ carrying the parametric curve $g(t)=(t^2,t)$ [for which the algebraic criterion holds, of course] to the parametric curve $f(t)=(t^2,t+t^3)$ [for which it fails]. To do this, let $\rho\colon\Bbb R\to [0,1]$ be a smooth function with $\rho(x) = 0$ for all $x\le -1/2$ and $\rho(1)=1$ for all $x\ge 0$. Define $F\colon\Bbb R^2\to\Bbb R^2$ by $$F(x,y) = \big(x,y(1+\rho(x)x)\big).$$ You can check that $F$ is a diffeomorphism (it's easy enough to write down the global inverse function and see it's smooth). And $F(g(t)) = F(t^2,t) = \big(t^2,t(1+\rho(t^2)t^2)\big) = (t^2,t+t^3) = f(t)$, as promised.

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