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I originally posted this question on the Mathematics StackExchange and got told to consider putting it on here, on MathOverflow. I will word the question a bit differently:

Let $X$ be a compact $k$-dimensional submanifold of $\mathbb{R}^n$, $Y\subseteq\mathbb{R}^n$ and $f\colon X\to Y$ be measurable. What do we need from $n$, $k$ and $Y$ in order to be able to find $T>0$ and some bounded $g\in C([0,T]\times\mathbb{R}^n, \mathbb{R}^n)$, $g$ locally Lipschitz w.r.t the second argument, such that if $u_{(g,x_0)}$ is the unique solution to the initial value problem $$ \dot{x}(t) = g(t,x(t)),\; t\in [0,T],\quad x(0) = x_0$$ we have $f \approx \left(f_g\colon X\to Y, x_0\mapsto u_{(g,x_0)}(T)\right)$ in some reasonable way?

E.g., is the set $\{f_g\colon\text{$g$ as above}\}$ dense in $(C(X,\mathbb{R}^n),\Vert\cdot\Vert_\infty)$? Is the answer trivially negative? What about the relationship between $k$ and $n$? Can someone point me to where to look for answers to this question? Does this question have a name?

Many thanks in advance. (Note that I am an undergraduate, so please excuse me and inform me if this question or its formulation are not relevant to this StackExchange)

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    $\begingroup$ Denote by $G(0,t)$ the solution operator of the ODE: $G(0,t)x_0$ is the value at time $t$ of the solution starting at $x_0\in X$. From your assumptions it follows that for any $t\in\mathbb{R}$, $G(0,t)$ is well-defined, and is a diffeomorphism of $X$. Further, $G(0,0)$ is just the identity. So, $[0,T]\ni t\mapsto G(0,t)$ provides an isotopy between identity and $G(0,T]$. And, if $X$ is orientable, isotopy preserves orientation. $\endgroup$ – user539887 Dec 12 '18 at 11:49
  • $\begingroup$ @user539887 Thank you. While I am still working on dissecting your comment, I wanted to let you know that I am not speaking of an ODE on the subset $X$, but of an ODE on $\mathbb{R}^n$; $X$ just collects the initial values I am interested in. So, I am not sure about $G(0,t)$ being a diffeomorphism of $X$ for a given $t$. Am I confusing something? $\endgroup$ – Ramen Dec 12 '18 at 15:08
  • $\begingroup$ I see. The image of $X$ under $G(0,T)$ is contained in $X$, and it appears that it is the whole of $X$, and the restriction $G(0,T)|_X$ is a diffeomorphism from $X$ into $X$. But we know only that $G(0,t)X$ is, for $t\in(0,t)$, a subset of $\mathbb{R}^n$ diffeomorphic to $X$. Surely, my isotopy argument breaks down. $\endgroup$ – user539887 Dec 12 '18 at 19:26
  • $\begingroup$ @user539887 I don't necessarily want $G(0,t)X\subseteq X$, too. Does it follow from my premises? I fail to see how. In fact, I hoped for some answer along the lines: "Provided $j\leq k<n$, $G(0,T) = f_g$ can approximate any continuous map $f\colon X\to\mathbb{R}^n$ for which $f(X)$ is subset of a compact $j$-dimensional submanifold of $\mathbb{R}^n$ (by varying $g$)". Would a such statement, in principle, even be compatible with my premises? Another thing, where could I read about the invertibility of $G(0,t)$ in $C^q$? Thanks again! $\endgroup$ – Ramen Dec 12 '18 at 21:30
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    $\begingroup$ Perhaps you need something like: "$G(0,T)X$ is contained in some preassigned set $Y$"? And the invertibility follows by the existence and continuous dependence theorem. $\endgroup$ – user539887 Dec 13 '18 at 6:03

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