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?
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?
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.