TITLE: Quadric in $n$ dimensions is a manifold?
QUESTION [5 upvotes]: I can show that the sphere $x_1^2+x_2^2+\ldots+x_n^2=1$ is an $(n-1)$-dimensional manifold by considering the map $f(x_1,\ldots,x_n)=x_1^2+\ldots+x_n^2$, and noticing that $1$ is a regular value of $f$, i.e. $Df(p)$ is surjective for all $p\in f^{-1}(1)$.
What about the quadric $x_1^2+x_2^2+\ldots+x_{n-1}^2=x_n^2$? I want to use the same trick, so I consider the map $g(x_1,\ldots,x_n)=x_1^2+\ldots+x_{n-1}^2-x_n^2$. Now, if $p=(p_1,\ldots,p_n)$, $Dg(p)$ is the $1\times n$ matrix $$\begin{bmatrix}2p_1 & 2p_2 & \ldots & -2p_n \end{bmatrix}.$$
Unfortunately, $Dg(p)$ is not surjective when $p=(0,0,\ldots,0)$.
How can I show that the quadric is a ($(n-1)$-dimensional?) manifold?

REPLY [3 votes]: The property of being a manifold is (if we assume second countability from the beginning) a local property: it is something that fails or holds on small scale, in a neighborhood of every point. Thus, one can speak  of manifold points of a set that is not a manifold itself. A manifold point is one that has a neighborhood homeomorphic to a Euclidean space. Or diffeomorphic, if we want to talk about smooth manifolds and have a notion of differentiability (which we do for submanifolds of a Euclidean space, as here). 
The cone $x_1^2+x_2^2+\ldots+x_{n-1}^2=x_n^2$ is not a manifold, only because of the vertex. Every other point is a manifold point, with a well-defined tangent. Another way to put it: cone minus its vertex is a manifold (as Daniel Fischer said), because the implicit function theorem applies to the defining function $x_1^2+x_2^2+\ldots+x_{n-1}^2-x_n^2$ on the set $\mathbb R^n\setminus\{0\}$.