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Let $B$ be an infinite-dimensional Banach space, and let $M\subset\mathbb{R}^n$ be a neighborhood of the origin in $\mathbb{R}^n$.

Suppose that $I:M\to B$ is a real-analytic function with $I(0)=0$ and such that the derivative of $I$ at $0$ has maximal rank.

Is it true that there exist neighborhoods $U,V\subset B$ of $0$ and a real-analytic diffeomorphism $\phi:U\to V$ such that $\phi\circ I$ is the restriction of a linear map $\mathbb{R}^n\to B$? If so, what is a good reference?

EDIT: I asked this question in the real-analytic setting, but might as well have done so in the complex-analytic case.

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I cannot give you a reference, but the answer ought to be yes.

To simplify notation, identify $\mathbb{R}^n$ with its image in $B$ under the derivative $I'(0)$. That image, being finite-dimensional, is the range $pB$ of a finite rank, bounded projection $p$. Replacing $M$ by a smaller neighbourhood if necessary, we can assume that $p\circ I$ is a diffeomorphism of $M$ onto a neighbourhood $N$ of $0$ in $pB$. Let $h\colon N\to M$ be its inverse, and define $\phi\colon p^{-1}(N)\to p^{-1}(M)$ by $$\phi(w)=x+w-I(x),\qquad x=h(pw).$$ If $w\in pB$ then $h(p(I(w)))=w$, so $\phi(I(w))=w+I(w)-I(w)=w$ – i.e., $\phi\circ I$ is the identity on $pB$.

To find the inverse of $\phi$, note that if $y=\phi(w)$ and $x=h(pw)\in pB$ then $py=x+pw-p(I(h(pw))=x$, so $w=y-x+I(x)=y-p(y)+I(py)$, i.e., $$\phi^{-1}(y)=y-py+I(py).$$

Edit: On second thought, it would have been more natural to define $\phi^{-1}$ first, with the requirement that its restriction to $pB$ be $I$. Letting its restriction to $(1-p)B$ be the identity is the simplest way to make it a diffeomorphism.

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Yes, this looks good. I was sort of hoping for a reference to the literature (as I'm essentially asking for a version of the constant rank theorem), so I will leave the question open for a while in case someone has a pointer! –  Lasse Rempe-Gillen May 1 '10 at 21:53

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