# Constant-Rank Level Set Theorem Proof

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 $$S \cap U$$?

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 $$\Phi(q)=c \iff \psi(\Phi(q))= (0,\ldots,0) \iff \psi\Phi\phi^{-1}(\phi(q)) = (0,\ldots,0) \iff \text{the first r coordinates of }\phi(q) \text{ are zero}$$