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I have a question about the rank theorem in Rudin's PMA book in page 229. Roughly speaking, it's a statement that for a continuous differentiable map (which can be linearly approximated), certain part of the map can be made precisely linear by a diffeomorphism of the domain of the map (and the map can be made fully linear (projection as a special case) if we are further allowed to do a diffeomorphism of the codomain).

The statement in Rudin roughly goes as follows. Let $F$ be a $C^1$ map from $\mathbb{R}^n$ to $\mathbb{R}^m$, with its derivative having constant rank $r$ in a neighborhood of a point $a$. Let $Y_1$ be the range of $F^{'}(a)=A$ (the image of the linear map I think), and let $P$ be a projection in $\mathbb{R}^m$ whose range is $Y_1$, $Y_2$ be the nullspace of $P$. Then there are open sets in $\mathbb{R}^n$, $U$ as a neighborhood of $a$ and $V$, a $C^1$ bijection (diffeomorphism) $H$ from $V$ to $U$, such that \begin{equation} F(H(x)) = A x + \varphi(A x) \end{equation} where $\varphi$ is a $C^1$ mapping of the open set $A(V) \in Y_1$ into $Y_2$.

My question is very simple. Since the map is linearly approximated as $F(x) \sim F(a) + F^{'}(a)(x-a) $. Shouldn't the statement be $F(H(x-a))-F(H(0)) = A (x-a) + \varphi(A (x-a))$ instead? (It's the small change of the function, not the value of the function itself, that can be made linear.)

liyiontheway
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If you write $F(H(0))$ as $Ax_0 + y_0$ for some $x_0 \in \mathbb{R}^n$ and some $y_0 \in \ker P$, and consider $\psi : t \in A(V + x_0) \mapsto \varphi(t - Ax_0) + y_0$, then $\psi$ is still $\mathcal{C}^1$ with its range in $\ker P$, and: $$F(H(x - a)) = A(x - a + x_0) + \psi(A(x - a + x_0))$$ Hence, when taking $\tilde{H} : x' \in V + x_0 \mapsto H(x' - x_0)$: $$F\left(\tilde{H}(x')\right) = Ax' + \psi(Ax')$$ Which means that you can just see your claim as a potential realisation of Rudin's theorem.

Bruno B
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