I am working on some properties of diffeomorphisms and came across the following question. For manifolds $M,N$, suppose I have for every $p \in M$ a linear map $F_p: T_p M \to T_x N$ for some $x \in N$.
Is there any characterization of when I can find a smooth map $\phi: M \to N$ such that $d\phi_p = F_p$ and $\phi(p) = x$?
My initial guess would be some sort of commutativity condition
$$ \begin{array}{ccc}
T_pM & \rightarrow & M \\
F_p \downarrow & & \phi \downarrow \\
T_x N & \rightarrow & N
\end{array} $$
However, I think this is not the answer since the map from $T_p M \to M$ is not clear.
Alternatively, perhaps there is a characterization of how to stitch together germs of smooth maps into a section on the sheaf of smooth maps from $M$ to $N$?
Best Answer
By assumption, we have for every $p \in M$ an $x \in N$ such that we have a linear map $F_p : T_p M \to T_x N$. We set $\phi(p) := x$. This defines a map $\phi : M \to N$. By gluing together the $F_p$ as $p$ varies over $M$, we get a map $F : TM \to TN$. By definition, the square $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccc} TM & \ra{F} & TN \\ \da{\pi_M} & & \da{\pi_N} \\ M & \ra{\phi} & N \end{array} $$ commutes. Note that $\phi$ is the unique map that makes this diagram commute.
If $\phi$ were smooth and $d\phi = F$, then $F$ would be smooth. So this is certainly a necessary condition for $\phi$ to be smooth. But it is also sufficient: if $F$ is smooth, then $\pi_N \circ F$ is smooth, and this map is constant on the fibers of $\pi_M$. Since $\pi_M$ is a submersion, this implies that $\phi$ is smooth.