Let $X,Y$ be $\mathbb R$-Hilbert spaces, $M\subseteq X$ be open, $\Phi\in C^1(M,Y)$ and $u\in M$ such that ${\rm D}\Phi(u)$ is surjective. Now let $X_0:=\ker{\rm D}\Phi(u)$.
Why is $X=X_0\oplus X_1$ for some closed $X_1\subseteq X$ and why is $A:=\left.{\rm D}\Phi(u)\right|_{X_1}$ bijective?
Clearly, the decomposition $X=X_0\oplus X_1$ would be possible using orthogonal decomposition in Hilbert spaces if $X_0$ would be closed. But unless ${\rm D}\Phi(u)$ is continuous, $X_0$ doesn't need to be closed. Is the decomposition still possible (maybe lacking the orthognality).
The second should be trivial, since $\left.{\rm D}\Phi(u)\right|_{X_0}=0\in AX_1$ or am I missing something?
Let $T:={\rm D}\Phi(u).$ Then, by definition(!), $T:X \to Y$ is a bounded linear operatot, hence $X_0$ is closed. With $X_1:= X_0^{\perp}$, we have $X=X_0\oplus X_1$.
Furthermore:
$A(X_1)=T(X_1)=T(X)=Y$ and $ker(A)=\{0\}.$