Suppose I have a $C^{r}$ diffeomorphism $F: \mathbb{R}^{n} \mapsto \mathbb{R}^{n}$ that I write as $x_{k+1} = F(x_{k})$, with $x \in \mathbb{R}^{n}$. Note that here $x_{k+1}$ is the image of $x_{k}$ under $F$ (i.e. we consider a dynamical system). Suppose I give a small $O(\epsilon)$ perturbation to $F$, and the resulting perturbed map is written in an implicit form as:
$$x_{k+1, \epsilon} = F(x_{k}) + \epsilon G(x_{k}, F(x_{k}), \epsilon)$$
for some $C^{r}$ function $G$ of both the preimage $x_{k}$, the image of $x_{k}$ under the unperturbed function $F$, and small parameter $\epsilon$.
How do I set up implicit function theorem to indeed verify that the perturbed map is again a $C^{r}$ diffeomorphism? Or do I use inverse function theorem? Also, on what does the upper bound on $\epsilon$ depend on, in order to have a diffeomorphism?
Edit. Can I simply define a new function $\Phi: \mathbb{R} \times \mathbb{R}^{n} \mapsto \mathbb{R}^{n}$ with $\Phi(\epsilon, x) \mapsto (F + \epsilon G)$ (here I suppressed arguments in $F$ and $G$) and notice that $D_{x}\Phi(0,x)$ is clearly invertible at all $x \in \mathbb{R}^{n}$ and then state that implicit function theorem says there is a unique $C^{r}$ function $x(\epsilon)$ defined for small $\epsilon$ ?
Another edit. I suppose I could apply inverse function theorem, but then I run into the problem of showing that a small linear perturbation of a linear isomorphism is also an isomorphism - is this true?