Let $f:I\subseteq \mathbb{R} \to \mathbb{R}^n$ be a smooth function ($I$ is an interval in $\mathbb{R}$). What are the sufficient conditions (or sufficient and necessary) under which the image of this function (say $S$ which is indeed a curve in $\mathbb{R}^n$) is a submanifold? Is there any famous theorem on this topic?
I tried to find such conditions in this way: according to the definition of submanifold, we need a 1-dimensional submanifold chart for $S$ at every point $x$ of $S$. Thus, there should exist open interval $W \subseteq \mathbb{R}$ and smooth function $G: W \to \mathbb{R}^n$. We need $G(W)=U \cap S$ where $U\subseteq \mathbb{R}^n$ is an open set and $x\in S \cap U$. Finally, we need the function $G$ to have the form $G(w)=A\left( {\begin{array}{*{20}{c}} w\\ g(w) \end{array}} \right)$ where $A$ is non-singular and $g:W\to\mathbb{R}^{n-1}$ is a smooth function (I used the definition from "Ordinary Differential Equations with Applications by C. Chicone").
I think the sufficient condition is that $f$ should be a diffeomorphism (it should be smooth and have smooth inverse) and $I$ be an open interval. Actually, in this case, the function $f$ plays the role of the function $G$ in the definition of submanifold.
Thank you, in advance, for your help!
Incidentally, I do not understand why we need $U$ in the definition of submanifold to be open? I am saying this because $S$ can be a closed subset of $\mathbb{R}^n$ and thus $U\cap S $ may not be open. Indeed, my question is that whay the image of $W$ under $G$ i.e., $G(W)=U \cap S$, is always open?
Thanks.
I guess you mean "embedded submanifolds" here? We say $M$ is an embedded submanifold of $N$ (where $M$ and $N$ are smooth manifolds) if some $f\colon M\to N$ is a smooth immersion and a topological embedding, i.e., if $df_p$ is nonsingular for each $p\in M$ and $f\colon M\to f(M)$ is a homeomorphism.
Now take $M=I$ (an interval in $\mathbb R$) and $N=\mathbb R^n$. In this case $df_p$ being nonsingular for each $p\in I$ means that $(\frac{df_1}{dt},\cdots,\frac{df_n}{dt})\neq 0$ at each $t\in I$, where $f=(f_1,\cdots,f_n)$ with $f_i\colon I\to \mathbb R$ smooth.
As for why you need open $U\subset \mathbb R^n$, the requirement that $f$ being a topological embedding addresses it. Note that if $f\colon I\to f(I)=S$ is a homeomorphism while $S$ inherits the topology from $\mathbb R^n$, then each open subset $V$ of $I$ is carried into an open subset $U\cap S$ of $S$, where $U\subset\mathbb R^n$ is open.
ps. If you let $f$ be a homeomorphism it's okay but this case forces $n=1$, which is trivial. I would recommend you to have a look over Frank W. Warner's Foundations of Differentiable Manifolds and Lie Groups, which is quite a good resource to learn differentiable manifolds quickly.