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

Question

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.

Answer 1

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.