# Traces of manifold-valued Sobolev maps

Let $$(M^m,g)$$ be a compact Riemannian manifold with smooth nonempty boundary, and $$N^n\subseteq \Bbb R^d$$ a boundaryless isometrically embedded Riemannian manifold. For $$1\le p<\infty$$ we define as usual $$W^{1,p}(M,N):=\{u\in W^{1,p}(M,\Bbb R^d):u(x)\in N\text{ a.e.}\}.$$ Using the Euclidean trace theorem, we have a trace map $$T_N:W^{1,p}(M,N)\to L^p(\partial M,\Bbb R^d)$$ just by restricting the domain of $$T:W^{1,p}(M,\Bbb R^d)\to L^p(\partial M,\Bbb R^d)$$. Is it true that the image of $$T_N$$ is contained in $$L^p(\partial M,N)$$?

If $$N$$ and $$p$$ are such that $$C^\infty(\overline M,N)$$ is not dense in $$W^{1,p}(M,N)$$, then I don't see how this can be proved. If we approximate $$u\in W^{1,p}(M,N)$$ by smooth functions $$u_j\in C^\infty(\overline M,\Bbb R^d)$$, then the continuity of the trace map implies $$Tu_j\to T_Nu$$ in $$L^p(M,\Bbb R^d)$$, so $$Tu_j\to T_Nu$$ a.e. in $$\partial M$$. But the images of the $$Tu_j$$'s are not necessarily in $$N$$, so this doesn't say anything about $$T_Nu$$. Now $$u_j\to u$$ a.e., but since $$\partial M$$ has measure zero we run into the precise issue that forces us to have a trace theorem in the first place!

In this article by Bethuel and Demengel, they say this is easy to see (top of the second page).