Intersection of the domains of the inverses of a one-parameter family Context.
Let $f_c:\mathbb{C}\rightarrow\mathbb{C}$ be a holomorphic function, depending holomorphically on the parameter $c\in\mathbb{C}$. Let $\alpha_0$ be a geometrically attracting fixed point of $f_{c_0}$, i.e. $f_{c_0}(\alpha_0) = \alpha_0$ and $\lambda_0 = f_{c_0}'(\alpha_0) \in \mathbb{D}\backslash\{0\}$.
Suppose there exists a neighbourhood $\Delta\ni c_0$ where we can define, using the implicit function theorem a holomorphic function $\alpha:\Delta\rightarrow\mathbb{C}$ such that $f_c(\alpha(c)) = \alpha(c)$ and naturally $\alpha(c_0) = c_0$, and additionally $\lambda(c) \equiv f_{c}'(\alpha(c))\in \mathbb{D}\backslash\{0\}$.
Now, for each $c\in\Delta$ there are going to be neighbourhoods $V(c)\supset U(c) \ni \alpha(c)$ where $f_c : V(c) \rightarrow U(c)$ is biholomorphic (i.e. it admits a holomorphic inverse).
Question.
Can we restrict $\Delta$ in such a way that
$$ \alpha(\Delta) \subset \bigcap_{c\in\Delta} V(c)$$
In other words, can we make $\Delta$ small enough so that for all $c_1,c_2\in\Delta$ we have $f_{c_2}(\alpha(c_1)) \in U(c_2)$?
Attempts.  I have managed to prove (using continuity) that I can make $\Delta$ small enough so that $\alpha(\Delta) \subset V_(c_0)$, but I am failing  to generalise to all parameters in $\Delta$. My intuition strongly says that this ought to be possible, but I fail to have a grasp of how "nicely" the domains $U(c)$ and $V(c)$ change. I tried to embed a disk around $\alpha(c)$ in $V(c)$, but I don't seem to get control on how fast it its radius changes. Maybe I need to use the implicit function theorem again?
(The context for this is that I want to generalise Koenig's Linearisation for a family of functions.)
Any help, advice or (partial) proof appreciated.  I apologise if this is obvious, but my background is Physics, and I am still learning.
 A: Lemma If $|f′(z)−f′(a)|<|f′(a)|$ for all $z$ in a disk containing $a$, then $f$ is invertible on that disk.
There is $r>0$ such that for all $z\in \mathbb{D}_{5r}(\alpha_0)$, 
$$ |f'_{c_0}(z) - f'_{c_0}(\alpha_0)| < \frac{1}{2} | f'_{c_0}(\alpha_0)| $$
We can then restrict $\Delta$ so that $\alpha(\Delta) \Subset \mathbb{D}_{r}(\alpha_0)$, which implies the following holds for all $c\in\Delta$:
$$\alpha(\Delta) \Subset \mathbb{D}_{2r}\big(\alpha(c)\big) \Subset \mathbb{D}_{4r}(\alpha_0)$$
Claim We can find a $\delta>0$ so that $\mathbb{D}_\delta(\alpha_0)\subset \Delta$  and all $c\in\mathbb{D}_\delta(\alpha_0) $ the $f_c$ are invertible on $\mathbb{D}_r(\alpha_0)$.
Proof  Note that the set
$$V = \bigcap_{c\in\alpha(\Delta)} \mathbb{D}_{2r}(\alpha(c))$$ is a subset of $\mathbb{D}_{4r}(\alpha_0) $ and as such, it is compactly contained in $\mathbb{D}_{5r}(\alpha_0)$.
Since the functions $f'_{c}$ converge to $f_{c_0}$ uniformly on compact subsets as $c\rightarrow c_0$, we can choose, for any $\epsilon >0$, a $\delta>0$ such that for all $c\in\mathbb{D}_\delta({c_0})$, and for all $z\in V$, 
$$|f'_c(z) - f'_{c_0}(z)| < \epsilon/2$$
 Additionally, we can make $\delta$ small enough so that we may assume $\mathbb{D}_\delta(c_0) \subset \Delta$ and
$$|f'_{c_0}(\alpha_0) - f'_c(\alpha(c))| < \epsilon/2 ~~~\mbox{ and }~~~ |f'_c(\alpha(c))| > \frac{1}{2}|f'_{c_0}(\alpha_0)| + \epsilon $$
For all $z\in \mathbb{D}_{2r}\big(\alpha(c)\big)$ and  $c\in\mathbb{D}_\delta(c_0)$, we then have $z\in V$ and
\begin{equation}
\begin{aligned}
|f'_c(z) - f'_c(\alpha(c))| 
&\leq |f'_c(z) - f'_{c_0}(z)| + |f'_{c_0}(z) - f'_{c_0}(\alpha_0)| + |f'_{c_0}(\alpha_0) - f'_c(\alpha(c))| \\
&< \frac{1}{2} | f'_{c_0}(\alpha_0)| +\epsilon \\
&<  |f'_c(\alpha(c))| 
\end{aligned}
\end{equation}
so that $f_c$ is invertible on $z\in \mathbb{D}_{2r}\big(\alpha(c)\big)$.  We conclude that each of the functions $f_c$ for $c\in \mathbb{D}_\delta(c_0)$ are invertible over $V \supset \alpha\big( \mathbb{D}_\delta(c_0)\big)$. Finally, it is easy to check that $\mathbb{D}_r(\alpha_0) \subset V$ $\square$.
