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$\newcommand{\R}{\mathbf R} \newcommand{\wh}{\widehat} \newcommand{\vp}{\varphi}$ I think the following should be true.

Let $S$ be an embedded $k$-submanifold of a smooth $n$-manifold $M$ and let $f:N\to TM$ be a smooth map defined on a smooth manifold $N$. Assume that $f(q)\in di(TS)$ for all $q\in N$, where $i:S\to M$ is the inclusion map. Then the unique map $F:N\to TS$ satisfying $f=di\circ F$ is also smooth.

Can anybody confirm if this is correct? Here is my argument:

Proof: Let $p$ be any point in $S$ and $(U, \vp)$ be a slice chart centered at $p$ for $S$ defined on a neighborhood $U$ of $p$ in $M$. Write $\vp(U)=\wh U$. Then we know that $\pi\circ \vp\circ i:U\cap S\to \pi(\wh U\cap (\R^k\times \mathbf 0))$ is a smooth chart on $S$ about $p$. We may assume that $\wh U$ is an open ball. Therefore, $\pi(\wh U\cap (\R^k\times \mathbf 0))=\pi(\wh U)$. The corresponding smooth chart $\Theta:T(U\cap S)\to \pi(\wh U)\times \R^k$ on $TS$ is defined as $$ \Theta(q;\ v)=(\pi(\vp(q)),\ d(\pi\circ \vp\circ i)_qv)=(\pi(\vp(q)),\ d(\pi\circ \vp)_qdi_qv) $$ for all $(q;\ v)\in T(U\cap S)$. From this it is easily checked that $\Theta\circ F(q)=(\pi\circ\vp(q),\ \pi\circ d\vp_qf(q))$ and therefore $\Theta\circ F$ is a smooth map. This shows that $F$ is locally smooth and hence $F$ is smooth.

caffeinemachine
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1 Answers1

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Yes it is correct. Here is another proof:

$di:TS\to TM$ is a smooth embedding, so $di(TS)$ is an embedded submanifold of $TM$ (see here for a proof of a more general statement). Hence, since $f(N)\subseteq di(TS)$, $f$ restricts to a smooth map $f:N\to di(TS)$ (by Corollary 5.30 in Lee's Introduction to Smooth Manifolds). Now, compose with the diffeomorphism $di(TS)\to TS$.

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Spenser
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