The following is a theorem from Fischer and Marsden's 1975's paper: Linearization stability of nonlinear PDEs.

Theorem.

Let $X, Y$ be Banach manifolds and $\Phi: X \rightarrow Y$ be $C^1$. Let $x_0 \in X$ and $\Phi\left(x_0\right)=y_0$. Suppose that $T \Phi\left(x_0\right)$ is surjective with the splitting kernel. Then the equation $\Phi(x)=y_0$ is linearization stable about $x_0$.

Proof. From the implicit function theorem, the set $\Phi^{-1}\left(y_0\right)$ is a $C^1$ submanifold near $x_0$ with tangent space the kernel of $T \Phi\left(x_0\right)$. Thus $h \in T_{x_0} X$ is a first order deformation iff $h \in \operatorname{ker} T \Phi\left(x_0\right)$ iff $h \in T_{x_1}\left(\Phi^{-1}\left(y_0\right)\right)$, and since $\Phi^{-1}\left(y_0\right)$ is a submanifold, there exists a curve $x(\lambda) \in \Phi^{-1}\left(y_0\right)$ which is actually tangent to $h$. (QED)

Can someone please explain the following sentence?
**From the implicit function theorem, the set $\Phi^{-1}\left(y_0\right)$ is a $C^1$ submanifold near $x_0$ with tangent space the kernel of $T \Phi\left(x_0\right)$.**

How the implicit function theorem (Banach space version) gives us $T_{x_{0}} \phi^{-1}(y_{0})$ = kernel $(T \phi (x_{0})$? The remaining part of the proof is straightforward.

Most importantly, what is splitting kernel in this context? I worked with some examples that showed that the surjectivity of the map $\phi$ is required; otherwise, we can create counterexamples. Please note that I posted this problem here, and I'm sorry if this problem is inappropriate for this site. Thanks so much.

Edit: Thanks again. I think now I understand the proof.

surjective with splitting kernelyou can also say that $T\phi(x_0)$ has acontinuous linear right inverse. $\endgroup$