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In baby Rudin, Rudin stated the rank theorem below:

9.32 Theorem

Suppose $m, n, r$ are nonnegative integers s.t. $m\geq r$, $n\geq r$. Let $F$ be a $C^{1}$ mapping of an open set $E\subseteq \mathbb{R}^{n}$ into $\mathbb{R}^{m}$, and $F'(x)$ has rank $r$ for every $x\in E$.

Fix $a\in E$, put $A = F'(a)$, Let $Y_{1} = \mathop{\mathrm{Im}}A$, let $Y_{2}$ be a complement space of $Y_{1}$ in $\mathbb{R}^{m}$, and let $P$ be the projection $\mathbb{R}^{m} = Y_{1}\oplus Y_{2}\to Y_{1}$.

Then there are open sets $U$ and $V$ in $\mathbb{R}^{n}$, with $a\in U$, $U\subseteq E$ and there is a 1-1 mapping $H$ of $V$ onto $U$(whose inverse is also of class $C^{1}$) such that

$$ F(H(x)) = Ax + \varphi(Ax)\ (x\in V) $$ where $\varphi$ is a $C^{1}$ mapping of open set $A(V)\subseteq Y_{1}$ into $Y_{2}$.

My Question is:

It seems to be $A(V)$ is not open in $\mathbb{R}^{m}$. Therefore the definition of differentiation of $\varphi\colon A(V)\to Y_{2}$ seems to be nonsense, so is the definition of $C^{1}$-mapping. Is there any justification?

K. Y.
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    Rudin is claiming that $A(V)$ is open in $Y_1$ (as it should be, because $A$ is a surjective linear mapping of $\Bbb{R}^n$ onto $Y_1$, hence it sends open subsets of $\Bbb{R}^n$ to open subsets of $Y_1$). Since $Y_1$ is a normed vector space (in fact a Banach space, since it is finite dimensional), the notion of differentiability makes perfect sense (you just have to modify definition 9.11 slightly see this for instance). Alternatively, you can just use an isomorphism to map $Y_1$ onto $\Bbb{R}^r$ and look at the corresponding function there. – peek-a-boo Jan 24 '21 at 10:03
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    i.e choose an isomorphism $\theta:Y_1\to\Bbb{R}^r$ and then define $\phi:A(V)\to Y_2$ to be $C^1$ if and only if $\phi\circ \theta^{-1}: \theta(\phi(A))\to Y_2\subset \Bbb{R}^n$ is $C^1$ in the "usual sense" (this definition is of course independent of $\theta$; and while this type of thinking is useful later on in differential geometry, I think the link above provides a clearer definition for now). – peek-a-boo Jan 24 '21 at 10:06
  • I will try it. Thanks. – K. Y. Jan 27 '21 at 08:44

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