Consider the following setup. Let $K$ be a compact topological space, $X$ a Fréchet space and $T:K \times X \to X$ a continuous family of linear maps (i.e. $T$ is a continuous map and $T\_k \equiv T(k, \cdot): X \to X$ is a linear for all all $k \in K$).


I would like to have a result similar to the following statement:



> 
> For every $k\_0$ with $T\_{k\_0} = \mathrm{id}\_X$, there exist an open neighborhood $U$ of $k\_0$ in $K$ such that $T\_k: X \to X$ is a topological isomorphism for every $k \in U$ and the map $T^{-1}: U \times X \to X$ defined by $(k, x) \mapsto T\_k^{-1}(x)$ is continuous.
> 
> 
> 


Since the space of invertible operators is not open in the space of all linear operators on a Fréchet space, such a result may be a bit much to ask but I was hoping that compactness of $K$ may help in this situation. I'm grateful for every pointers to results that go into this direction.



Openness of invertibility in Fréchet spaces for families parameterized by compact spaces
The statement is indeed overoptimistic.


As a counterexample, consider the continuous family $T: [0,1] \times C^\infty([0, 1]) \to C^\infty([0, 1])$ of linear differential operators defined by
 \begin{equation}
 T\_r (g) (x) = g - r x g'
 \end{equation}
 for $r \in [0,1]$ and $ g \in C^\infty([0, 1]) $.
 Clearly, $ T\_0 $ is the identity operator on $ C^\infty([0, 1]) $.
 On the other hand, for every $ n \in \mathbb{N} $, the operator $ T\_{\frac{1}{n}} $ annihilates the function $ g\_n(x) = x^n $.
 Hence, $ T\_r $ fails to be injective for arbitrarily small $ r $.


