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This is a proof from Lee's Introduction To Smooth Manifolds. I don't understand how the red part follows. The Rank Theorem will tell us that on $U$, $\Phi$ has coordinate representation:

$$\hat{\Phi}(x^1,\ldots,x^r,x^{r+1},\ldots,x^n) = (x^1,\ldots,x^r,0,\ldots,0)$$

I suppose something special happens in particular on $S \cap U$ which is that slice set mentioned. Why do $x^{r+1},\ldots,x^n$ switch from being $0$ to non-zero while $x^1,\ldots,x^r$ switch from being non-zero to $0$ on $S \cap U$?

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1 Answer 1

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This just has to do with the charts $(U,\phi)$, and $(V,\psi)$ being "centered" at $p$ and $c$, i.e., $\phi(p)=\vec{0}\in \mathbb{R^m}$ and $\psi(c)=\vec{0}\in \mathbb{R}^n$.

By rank theorem, we have $\psi \circ \Phi \circ \phi^{-1}: (x^1,\ldots,x^r,x^{r+1},\ldots,x^m)\mapsto (x^1,\ldots,x^r,0,\ldots,0) $ as you said. And if we want a point $q\in M$ to map specifically to $c$, we should have $$ \Phi(q)=c \iff \psi(\Phi(q))= (0,\ldots,0) \iff \psi\Phi\phi^{-1}(\phi(q)) = (0,\ldots,0) \iff \text{the first r coordinates of }\phi(q) \text{ are zero} $$

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