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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?

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    $\begingroup$ 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. $\endgroup$ Commented Jan 30, 2019 at 1:08
  • $\begingroup$ thanks for the reference!! $\endgroup$
    – jason
    Commented Jan 30, 2019 at 3:05

1 Answer 1

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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.

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  • $\begingroup$ yes, Thanks for your answer. $\endgroup$
    – jason
    Commented Jan 30, 2019 at 3:05
  • $\begingroup$ @jason You are welcome! $\endgroup$ Commented Jan 30, 2019 at 3:08
  • $\begingroup$ @ Piotr Hajlasz to prove one-one, can we just do it in this way without embedding thm: let $d$ is metric, $d(f_{\epsilon}(x), f_{\epsilon}(y)) \ge \inf \det (f_{\epsilon}) \cdot d(x,y)$? $\endgroup$
    – jason
    Commented Jan 30, 2019 at 18:59
  • $\begingroup$ @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. $\endgroup$ Commented Jan 30, 2019 at 19:50

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