# Can a smooth map between two embedded submanifolds be (locally) smoothly extended?

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 $$\bar F \colon \cal E \to \cal E'$$?

• Do you want a map defined on all of $\mathcal E$ or will a neighborhood $U$ of a point in $M$ do? Remember that maps to $\mathcal E'$ are vector-valued functions. – Ted Shifrin Jan 24 at 23:15
• I expect a smooth extension to all of $\cal E$ won't be possible, because already for $\cal M' = \mathbb{R}$ this requires $\cal M$ to be properly embedded (Lee, Lemma 5.34 and Exercise 5-18). So a smooth extension to a neighborhood of $\cal M$ in $\cal E$ will have to do. My difficulty is with the co-domain $\cal M' \subseteq \cal E'$. Would it suffice to simply consider $F$ as a map into $\cal E' \approxeq \mathbb{R}^d$, and to study its individual components? – Nicolas Boumal Jan 24 at 23:32
• (Also, I suspect that if such smooth extensions exist, they would exist regardless of the linear structure of $\cal E'$.) – Nicolas Boumal Jan 24 at 23:33
• Closely related: math.stackexchange.com/questions/1893383/… – Eric Wofsey Jan 25 at 6:31

## 1 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 $$M'$$).

• This is great, thanks Stephen! Further composing the map you propose with the inclusion map from M' into E' gives the final answer as a map from (a neighborhood of M in) E to E', and illustrates the role of having M' embedded in E' (so that the inclusion would be smooth). – Nicolas Boumal Jan 25 at 16:15
• Yes, good catch! – Stephen M Jan 25 at 19:39