In the following, all subsets of Euclidean spaces are considered with the respective subspace topologies. Define an equivalence relation \( \sim \) on the sphere \[ S = \left\{ (x_1, x_2, x_3) \in \mathbb{R}^3 : x_1^2 + x_2^2 + x_3^2 = 1 \right\} \] by \( (x_1, x_2, x_3) \sim (y_1, y_2, y_3) \) if \( x_3 = y_3 \), for \( (x_1, x_2, x_3), (y_1, y_2, y_3) \in S \). Let \( [x_1, x_2, x_3] \) denote the equivalence class of \( (x_1, x_2, x_3) \), and let \( X \) denote the set of all such equivalence classes. Let \( L : S \to X \) be given by \[ L\left( (x_1, x_2, x_3) \right) = [x_1, x_2, x_3]. \] If \( X \) is provided with the quotient topology induced by the map \( L \), then which one of the following is TRUE?

Question

Consider the following subsets of the Euclidean space \( \mathbb{R}^4 \):
\( S = \{ (x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : x_1^2 + x_2^2 + x_3^2 - x_4^2 = 0 \} \),
\( T = \{ (x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : x_1^2 + x_2^2 + x_3^2 - x_4^2 = 1 \} \),
\( U = \{ (x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : x_1^2 + x_2^2 + x_3^2 - x_4^2 = -1 \} \).
Then, which one of the following is TRUE?

Answer 1

To solve this problem, we need to understand the concept of quotient topology and the equivalence relation defined on the sphere \( S \).

The sphere \( S \) is defined as:

S = \left\{ (x_1, x_2, x_3) \in \mathbb{R}^3 : x_1^2 + x_2^2 + x_3^2 = 1 \right\}

An equivalence relation \( \sim \) is defined on \( S \) as \( (x_1, x_2, x_3) \sim (y_1, y_2, y_3) \) if \( x_3 = y_3 \). This means all points on the same horizontal plane (constant \( x_3 \)-coordinate) are equivalent.

The equivalence class \([x_1, x_2, x_3]\) consists of all points of the form \((x_1', x_2', x_3)\) for which \((x_1')^2 + (x_2')^2 = 1 - x_3^2\). Hence, each equivalence class is a circle in the plane of constant \( x_3 \), and its radius is given by \( \sqrt{1-x_3^2} \).

The set \( X \), comprised of equivalence classes, is indexed by the coordinate \( x_3 \). Thus, \( X \) is essentially the set of values \( x_3 \) can take, which is the interval \([-1, 1]\).

The map \( L \) defined by:

L\left( (x_1, x_2, x_3) \right) = [x_1, x_2, x_3]

projects each point on the sphere to its equivalence class, which corresponds to a specific value of \( x_3 \). This map is surjective, and in the quotient topology, \( X \) is homeomorphic to the space of distinct equivalence classes, which is the interval \([-1, 1]\).

Thus, the correct option is:

\( X \) is homeomorphic to \( \{ x \in \mathbb{R} : -1 \leq x \leq 1 \} \)

This conclusion follows because identifying points based on \( x_3 \) collapses each circle at \( x_3 = \text{constant} \) to a point identified by that \( x_3 \), resulting in the interval \([-1, 1]\) reflecting all possible values of \( x_3 \).

Answer 2