Proving $\det \big(Df\big|_x\big)=0$ for a function into unit circle Let $f:\mathbb{R}^2\to S$ where $S=\{x\in\mathbb{R}^2:\, ||x||=1\}$. Prove that $\det \big(Df\big|_x\big)=0$ for all $x$.
I'm having trouble attacking this. So I need to show that there is some $\mathbf{h}\neq 0$ such that $Df\big|_x(\mathbf{h})=0$, but I'm having trouble interpreting the end result and relating it to the rather vague conditions. One thought was to consider how $f$ acts on $S$ in the first place, by looking at $f\circ f$. Then:
$$\det  \big(Df\circ f\big|_x\big)=\det \big(Df\big|_{f(x)}\big)\det \big(Df\big|_x\big)$$
But then I'd need to argue that the LHS is $0$ and that the other RHS is not, which seems equally hard. Any ideas?
 A: The plane is a smooth two dimensional manifold and the circle is a smooth one dimensional manifold. The differential $Df|_x$ is a linear map between the tangent space to the plane at $x$, and the tangent space to the circle at $f(x)$. For each $x $, $Df|_x$ is a linear map between a two dimensional vector space and a one dimensional vector space.
The rank-nullity theorem tells us that if $L : U \to V$ is a linear map then 
$$\dim(\ker L) +\dim(\mathrm{im} \, L) = \dim U$$
In the case of $Df|_x : T_x\mathbb R^2 \to T_{f(x)}S$ we see that
$$ \dim(\ker Df|_x) + \dim(\mathrm{im}\, Df|_x) = 2$$
Since $0 \le \dim(\mathrm{im}\, Df|_x) \le 1$, it follows that $1 \le \dim(\ker Df|_x) \le 2$.
A: I'll write points in $\mathbb R^2$ as $z=(x,y).$ Let $f= (g,h).$ Then $g^2 + h^2 \equiv 1.$ Differentiating with respect to $x,y$ gives $2gg_x + 2hh_x = 0 = 2gg_y + 2hh_y.$ Thus $2g(g_x,g_y) + 2h(h_x,h_y) = (0,0)$ for every $z.$ Now $g,h$ cannot simultaneously vanish. So we'ver shown $(g_x,g_y), (h_x,h_y)$ are linearly dependent at every $z.$ Thus the determinant of the matrix whose rows are $(g_x,g_y), (h_x,h_y)$ is $0$ for all $z.$ But that determinant is exactly $\det Df(z).$
