Exercise 11, § 2

General Topology, VI

Let $C_{n, p}$ be the "quadric cone" in $\mathbf{R}^n$ defined by the equation

x_1^2 + x_2^2 + \cdots + x_p^2 - x_{p+1}^2 - \cdots - x_n^2 = 0 \quad (1 \leq p \leq n-1).

Show that the complement of

\{0\}

C_{n, p}

is homeomorphic to

S_{p-1} \times S_{n-p-1} \times \mathbf{R}

Solution

No solution has been written yet.