Let $S\subset M$ be a properly embedded submanifold (in particular, $S\subset M$ is a closed submanifold) and let $V$ be a smooth vector field on $M$ tangent to $S$. Since we have that $M\setminus S\subset M$ is open, we know that it is a submanifold of $M$ using the subspace charts. Does this imply that $V$ is tangent to $M\setminus S$?

Any ideas?

Edit: Here's an idea I had. $V$ is a smooth vector field on $M$, then $V_p\in T_pM$ for each $p\in M$. Then since $M\setminus S\subset M$ is open, the map $d\iota_p^{-1}:T_pM\to T_p(M\setminus S)$ is an isomorphism for $\iota:M\setminus S\to M$ the inclusion, so $d\iota_p^{-1}(V_p)\in T_p(M\setminus S)$. Then consider $d\iota^{-1}(V)$. Then this is a vector field (?) which is $\iota$ related to $V$, so $V$ is tangent to $M\setminus S$.

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    $\begingroup$ If $U$ is an open subset of $M$, then every vector field on $M$ is tangent to $U$, because $T_pU=T_pM$ for every $p\in U$. $\endgroup$ – Jack Lee Mar 9 '15 at 14:08
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    $\begingroup$ Right, that was much easier than I was making that out to be. Thank you and thank you for your wonderful text (which is what I'm studying from now). $\endgroup$ – Moya Mar 10 '15 at 0:33
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    $\begingroup$ You're welcome! $\endgroup$ – Jack Lee Mar 10 '15 at 0:38

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