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I'm having difficulty proving a well-known result from functional analysis. Any hints would be greatly appreciated.

Fix a Fréchet differentiable map of Banach spaces $g: X \to B$. Assume that, at a point $x_0 \in X$ where $g(x_0) = 0$, the derivative $Dg$ is Fredholm and surjective, so in particular the index of $Dg$ is the dimension of its kernel. A typical place this might arise is in solving PDEs, where $g$ is given by some nice differential operator and $x_0$ is some generic solution to the PDE.

I want to prove:

The moduli space $\{x \in X \, | \, g(x) = 0\}$ has a smooth chart near $x_0$ of dimension $index(Dg)$.

I know this is supposed to be a consequence of the implicit function theorem for Frechet maps, which I understand. I also understand the proof of the submersion theorem for finite-dimensional manifolds, which is simply an application of the implicit function theorem in finite dimensions. I am trying, in essence, to prove the above result the same way one would prove the finite-dimensional submersion theorem.

However, I'm having difficulty proving a version of the submersion theorem in the infinite-dimensional setting. I'll describe my approach here -- I suspect that maybe I'm trying to use the implicit function theorem in the wrong way.

Defining the finite-dimensional vector space $Y = kernel(Dg)$ and the Banach space $Z = B \times Y$, if one can choose a splitting map $\pi: X \to Y$, one can define a map $$F: X \times Y \to Z$$ given by $$F(x,y) = (g(x), \pi(x) - y).$$ Then the implicit function theorem obviously gives a parametrization of the set $\{x \, | \, g(x) = 0\}$ near $x_0$, in terms of an open set in $Y$.

The splitting map $\pi$ always exists when $X, B$ are finite dimensional, but my question is: can such a splitting always be chosen when $X, B$ are infinite-dimensional? And if not, how can I prove the main result above? I'm trying to follow the finite-dimensional analogies, perhaps too naively, and failing. Any help would be appreciated.

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If I understand correctly, you are asking why $Y$ is complemented in $X$. Every finite-dimensional subspace is complemented. Just write the identity map on $Y$ as a sum of rank-1 linear maps (essentially linear functionals) and extend them via Hahn-Banach. –  user53153 Dec 28 '12 at 19:23
    
Perfect! Thank you very much. –  user54535 Dec 29 '12 at 0:47
    
Glad to help. I'll copy the comment into the answer box, so that the question does not look unanswered. –  user53153 Dec 29 '12 at 0:49
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You are asking why $Y$ is complemented in $X$. Every finite-dimensional subspace is complemented. Just write the identity map on $Y$ as a sum of rank-$1$ linear maps (essentially linear functionals) and extend them via the Hahn-Banach theorem.

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