# How do you show that the class of diffeomorphisms from a compact manifold is stable?

This is pretty much a copy of this question. However, it is unanswered, which makes my question not a duplicate according to this meta answer. I need to show that the class of diffeomorphisms from $$N$$ to $$M$$ is stable, where $$N$$ is compact. I.e., I need to show that for a smooth family of functions $$\{F_s : N \longrightarrow M\}_{s\in S}$$ with $$F_0$$ a diffeomorphism, there is a neighborhood $$T \ni 0$$ such that $$F_s$$ is a diffeomorphism whenever $$s \in T.$$ This is part of Problem 6-16 from Lee's Introduction to Smooth Manifolds.

So far, I can show that $$F_s$$ is injective and a local diffeomorphism. If I could show it is surjective, then a proposition from Lee would give me diffeomorphism. But I can't think of how to do this. As an alternative, since $$N$$ is compact, I thought of getting a finite collection of domains on which $$F_s$$ restricts to a diffeomorphism and using a partition of unity to put them together, but the functions of a partition of unity are only guaranteed to be smooth, not diffeomorphisms, so the resulting composition would not necessarily be a diffeomorphism. I also considered using the fact that $$dF_0$$ is an isomorphism, but I got stuck here. I think that trying to show surjectivity sounds most promising, unless someone knows how to make the partition of unity work. Any help is appreciated.

So far, you have a connected neighborhood $$U\in S$$ such that $$F_s$$ is an injective local diffeomorphism for $$s\in U$$. You can show that these $$F_s$$ are also surjective for entirely topological reasons. Throughout, fix $$s\in U$$
$$F_s$$ is an open map, since it is a local diffeomorphism, and a closed map since $$N$$ is compact. Thus, if $$C\subseteq N$$ is a connected component, then $$F(C)$$ is as well. Additionally, $$F_s$$ must map connected components of $$N$$ to connected components of $$M$$ in the same way as $$F_0$$, since $$U$$ is connected, and thus we can find a path from $$F_0(x)$$ to $$F_s(x)$$ for any $$x\in M$$. Since $$F_0$$ is surjective, $$F_s$$ is as well.