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Let $M$ and $N$ be smooth manifolds, let $p \in M$ and let $L:T_p M \rightarrow T_{L(p)} N$ be a linear map.

I wonder if there always exists a smooth map $F:M \rightarrow N$ on a neighborhood of $p$ such that $L$ is the differential map of $F$ at $p$?

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  • $\begingroup$ Your question is worded a bit ambiguously, by suggesting the existence of this particular differential map $dF_p: T_p M \rightarrow T_{F(p)} N$, you inexplicitly suggest that $F$ is already a map (since you used $F(p)$ to parametrise N $\endgroup$ – Alon Yariv Sep 25 '19 at 9:47
  • $\begingroup$ I modified to question according your observation. $\endgroup$ – Juan Alvarado Sep 25 '19 at 10:47
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I think yes, as long as the discussion is only at a single point $p$. I think you mean $L:T_p M\to T_q N$ for some points $p$ and $q$, since $L(p)$ doesn't make sense, unless I've misunderstood.

Using charts which send $p$ and $q$ to 0, your question becomes "can I find a map between euclidean spaces, the derivative of which is the linear map $L$?" Since the derivative of a linear map is itself, you can take $f = L: \mathbb R^m \to \mathbb R^n$ (where you use the same chart coordinates to identify the tangent spaces with euclidean spaces) to be the representation of your map in these charts.

Then you can multiply by a bump function which is identically 1 on a smaller neighborhood of $p$ (so as not to affect the derivative at that point), in order to define a function on the whole of $M$.

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yes, suppose $dim M=m, dimN=n$. Consider charts $U$ and $V$ of $p$ and $f(p)$ that you identify to open neighborhoods of $0$ in $\mathbb{R}^m$ and in $\mathbb{R}^n$ with this identification $p$ and $f(p)$ are identified to $0$. Let $g$ be a cut off function defined on $V$: there exist neighborhoods $W\subset W_1\subset V$ of $0$ such that the restriction of $g$ to $W$ is $1$ and $g=0$ on $V-W_1$. Write $h=gf$.

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