How to prove $H^1(M) \subset H^s(M)$ is a continuous embedding for manifold $M$? Let $M$ be a $C^k$ manifold for some integer $k$. How does one show that
$$H^1(M) \subset H^s(M)$$
is continuous, where $s \in (0,1)$?
I was planning to pull back the norms onto a subset $D_i$ of Euclidean space via chart maps, and apply the result for open sets in Euclidean space and transfer back. But my problem is that the open set the chart map pulls back to, $D_i$, I need it to be at least a Lipschitz domain to apply the result for Euclidean space. How do I ensure that it is Lipschitz? AFAIK I know nothing about the $D_i$. 
Or is it the case that a $C^k$ manifold implies that the chart maps are actually $C^k-$diffeomorphism? In this case, maybe we can say that the chart maps map from an open ball around the boundary onto $D_i$, and because it's a diffeomorphism, perhaps $D_i$ also shares the same smoothness as the open ball?
 A: If $M$ is $C^k$, $k\ge 1$, then the charts are also $C^k$. Thus you can prove your inclusion as you suggest.
A: It mostly depends on how your Sobolev spaces are defined on $M$ in the first place.
That said, there are two ways around the potentially non-smooth edges of $D_i$.. 
a) You can use charts for which $D_i$ are open balls. Indeed, for every $x\in M$ there is a chart map $\varphi:D\to U$ from  open set $D$ onto a neighborhood $U$ of $x$. Let $B\subset D$ by an open ball containing $\varphi^{-1}(x)$. The restriction of $\varphi$ onto $ B $ is a new chart map $\tilde \varphi$.  Notice that $M$ is still covered by these new charts. 
Due to the above possibility, manifolds are often defined so that chart maps have an open ball, or the entire space $\mathbb R^n$ as the domain. 
b) Use a partition of unity $\chi_i$ subordinated to the cover of $M$ by charts. The function $u\in H^1(M)$ that you work with is the sum of $u\chi_i$. Each $u\chi_i$, when pulled to $\mathbb R^n$, has compact support in $D_i$, so the geometry of $\partial D_i$ is irrelevant. 
The second way is preferable because the compactness of support also gives uniform control on the derivative of a smooth map.
