Consider $\cal M$ and $\cal M'$, smooth embedded submanifolds of two linear manifolds $\cal E$ and $\cal E'$ (respectively). Let $F \colon \cal M \to \cal M'$ be a smooth map.
From Lee's textbook (2012, Intro to smooth manifolds), Lemma 5.34, we know that for the special case where $\cal M' = \mathbb{R}$ we can smoothly extend $F$, at least locally:
There exists an open neighborhood $U$ of $\cal M$ in $\cal E$ and a smooth function $\bar F \colon U \to \mathbb{R}$ such that $F$ is the restriction of $\bar F$ to $\cal M$, that is, $F = \bar F|_{\cal M}$.
My question: for the more general case where $\cal M'$ is not simply equal to $\mathbb{R}$ but can be any embedded submanifold of a linear manifold $\cal E'$, can I also have such a smooth extension with a map $\bar F \colon \cal E \to \cal E'$?
Best Answer
You should be able to do this using the tubular neighborhood theorem of Riemannian geometry. Let $\pi:NM \to M$ be the basepoint map from the normal bundle of M to M, and let $V$ be a sufficiently small neighborhood of (the canonical image of) $M$ in $NM$, such that there is a diffeomorphism $\psi: V \to U$ onto some tubular neighborhood of $M$ (by the TNT). Then set $\widetilde{F} = F \circ \pi \circ \psi^{-1}: U \to M'$. This is your extension (and it takes its values in $M'$).