# $C^1$ perturbation of diffeomorphism is diffeomorphism?

if $$f \in$$ diff($$M$$), where $$M$$ is manifold, if $$C^1$$ perturbation $$f_{\epsilon}$$ of $$f$$ s.t. $$||f_{\epsilon}-f||_{C^1} < \epsilon$$.

Can we prove $$f_{\epsilon} \in$$ diff($$M$$) if $$\epsilon$$ is small enough?

• Yes, $C^1$ diffeomorphisms form an open subset of the set of smooth self-maps of $M$ equipped with the strong $C^1$ topology, which equals to $C^1$ topology if $M$ is compact, see e.g. Hirsch, Differential topology, Chapter 2, theorem 1.6. Commented Jan 30, 2019 at 1:08
• thanks for the reference!! Commented Jan 30, 2019 at 3:05

Assuming that $$M$$ is a compact manifold, the answer is yes. Indeed, $$\det Df(x)\neq 0$$ for $$x\in M$$ and if $$|Df(x)-Df_\epsilon(x)|$$ is small, then $$\det Df_\epsilon(x)\neq 0$$, because the set of invertible matrices is open. Therefore $$f_\epsilon$$ is a local diffeomorphism. It remains to show that $$f_\epsilon$$ is one-to-one if $$\epsilon$$ is small.
Assume that $$M$$ is embedded into the Euclidean space (it is always possible to have a smooth embedding by Whitney theorem). Then the Riemannian metric is bi-Lipschitz equivalent to the Euclidean distance $$|x-y|$$.
Since $$M$$ is compact, there is $$C>0$$ such that $$|f(x)-f(y)|\geq C|x-y|$$ for all $$x,y\in M$$ and there is $$\epsilon>0$$ such that $$|(f_\epsilon-f)(x)-(f_\epsilon-f)(y)|\leq \frac{C}{2}|x-y|$$, for all $$x,y\in M$$, because the derivative of $$f_\epsilon-f$$ is small. Therefore $$|f_\epsilon(x)-f_\epsilon(y)|\geq |f(x)-f(y)|- |(f_\epsilon-f)(x)-(f_\epsilon-f)(y)|\geq \frac{C}{2}|x-y|,$$ proving that $$f_\epsilon$$ is one-to-one.
• @ Piotr Hajlasz to prove one-one, can we just do it in this way without embedding thm: let $d$ is metric, $d(f_{\epsilon}(x), f_{\epsilon}(y)) \ge \inf \det (f_{\epsilon}) \cdot d(x,y)$? Commented Jan 30, 2019 at 18:59
• @jason Embedding is not needed, but the inequality needs to be different. It suffices to prove the lower estimate . Instead of $\det$ you need to consider the infimum of smallest singular values of $Df(x)$. Writing everything with details to make the proof very rigorous could be annoying, but the idea is as explained in my answer. Commented Jan 30, 2019 at 19:50