Proposition

$\mathbb{R}^{n - 1}$ is homeomorphic to $\partial \mathbb{H}^n$.

Solution

Let $f: \mathbb{R}^{n - 1} \rightarrow \partial \mathbb{H}^n$ be defined such that $f(x_1, \cdots, x_{n - 1}) = (x_1, \cdots, x_{n - 1}, 0)$.

• Injective?
  • $f(x) = f(y)$ implies that $(x, 0) = (y, 0)$, so $x = y$.
• Surjective?
  • For any $(x_1, \cdots, x_{n - 1}, 0 \in \mathbb{H}^n$, $f(x_1, \cdots, x_{n - 1}) = (x_1, \cdots, x_{n - 1}, 0)$.
• Continuous?
  • Each $(x_1, \cdots, x_{n - 1}) \mapsto x_i$ is a canonical projection, so it is continuous. $(x_1, \cdots, x_{n - 1}) \mapsto 0$ is a constant function, so it is continuous. Since each component function of $f$ is continuous, $f$ is continuous.
• Open?
  • Let $U \subset \mathbb{R}^{n - 1}$ be open. Then $f(U) = U \times \{ 0 \} = (U \times \mathbb{R}) \cap \partial \mathbb{H}^n$. $U \times \mathbb{R}$ is open in $\mathbb{R}^n$, so $(U \times \mathbb{R}) \cap \partial \mathbb{H}^n$ is open in $\partial \mathbb{H}^n$.

Therefore, $f$ is a homeomorphism.