Let $f: S^{2} \rightarrow \mathbb{R}$ be the function defined by $f(x, y, z)=z$ (a) Recall that the rank rk $T$ of a linear transformation $T$ is the dimension of its image. For any $p \in S^{2}$, the image of $d f_{p}: T_{p} S^{2} \rightarrow \mathbb{R}$ is a subspace of $\mathbb{R}$. Thus its dimension is either 0 or 1 , so rk $d f_{p}$ is either 0 or 1 . For which $p \in S^{2}$ is rk $d f_{p}$ equal to $0,$ and for which $p \in S^{2}$ is it equal to $1 ?$ (b) For which $a \in[-1,1]$ is the preimage $f^{-1}(a) \subset S^{2}$ a smooth curve? (c) Describe a relationship between the answers to parts (a) and (b).
Let $X \subset \mathbb{R}^{n}$ be a $k$ -manifold, $Y \subset \mathbb{R}^{m}$ an $\ell$ -manifold, $f: X \rightarrow Y$ a smooth map, and $p \in X$ a point. Recall that $d f_{p}$ was defined as follows in lecture. First, we pick a local parametrization $\phi$ of $X$ near $p$ and a local parametrization $\psi$ of $Y$ near $f(p) .$ That is, $\phi: V \rightarrow U$ is a diffeomorphism from an open subset $V$ of $\mathbb{R}^{k}$ to an open subset $U$ of $X$ which contains $p$, and $\psi: V^{\prime} \rightarrow U^{\prime}$ is a diffeomorphism from an open subset $V^{\prime}$ of $\mathbb{R}^{\prime}$ to an open subset $U^{\prime}$ of $Y$ which contains $f(p)$. As discussed, we could assume that $f(U) \subset U^{\prime}$ and (for convenience) that $\phi(0)=p$ and $\psi(0)=f(p)$. Then $d f_{p}$ is the unique linear transformation such that $$ d f_{p} \circ d \phi_{0}=d \psi_{0} \circ d h_{0} $$ where $h=\psi^{-1} \circ f \circ \phi$ Since $f$ is smooth, there exists an open subset $W$ of $\mathbb{R}^{n}$ containing $p$ and a smooth map $F: W \rightarrow \mathbb{R}^{m}$ such that $\left.F\right|_{W \cap X}=f \mid W \cap X .$ We may assume that $U \subseteq W:$ if this were not the case, we could replace $U$ with $U \cap W, V$ with $\phi^{-1}(U \cap W),$ and $\phi$ with its restriction to $\phi^{-1}(U \cap W)$. Show that for all $v \in T_{p} X,$ we have $d f_{p}(v)=d F_{p}(v) .$ (Hint: look for the right way to use the chain rule.)