Are the spaces $\mathbb{H}^n$ and $\overline{\mathbb{R}}^n_+$ homeomorphic?

Question

Let $\mathbb{H}^n = \{(x^1, \dots , x^n) \in \mathbb{R}^n : x^n \ge 0\}$, and $\overline{\mathbb{R}}^n_+ = \{(x^1, \dots , x^n) \in \mathbb{R}^n : x^1 \ge 0, \dots, x^n \ge 0\}$. Endow each set with the subspace topology it inherits from $\mathbb{R}^n$.

Exercise 16.18 on page 415 of Lee's Introduction to Smooth Manifolds (2nd edition) asks us to show that $\mathbb{H}^n$ and $\overline{\mathbb{R}}^n_+$ are homeomorphic.

However, it seems to me the spaces are not homeomorphic, as shown by the following argument.

Suppose $f: \overline{\mathbb{R}}^n_+ \to \mathbb{H}^n$ is a homeomorphism. Set $f(0, \dots , 0) = (a^1, \dots, a^n)$. Then

$$A = f(\overline{\mathbb{R}}^n_+ \setminus \{(0, \dots, 0)\}) = f((\infty,0) \times \cdots \times (\infty,0)) = \\ ((\infty, a^1) \cup (a^1, - \infty)) \times \cdots \times ((\infty, a^n) \cup (a^n, 0]),$$ where, if $a^n= 0$, then in the last factor we discard $(a^n, 0]$.

The set $A$ is connected as it is the continuous image of the connected set $(\infty, 0) \times (\infty,0)$. But at the same time we see that $A$ is disconnected as it is the union of the disjoint relatively open sets $(\infty, a^1) \times \cdots \times (\infty, a^n)$ and $ (a^1, - \infty) \times \cdots \times (a^n, 0]$ (or $(\infty, a^1) \times \cdots \times (\infty, 0)$ and $(a^1, - \infty) \times \cdots \times (\infty, 0)$ in case $a^n = 0$).

Is there a mistake in my argument or something I am missing? Any comments are greatly appreciated.

Answer 1

We have Let $\mathbb{H}^n = \mathbb{R}^{n-1} \times [0,\infty)$ and $\overline{\mathbb{R}}^n_+ = [0,\infty)^n$. Let us first construct a homeomorphism $h_2 : [0,\infty)^2 \to \mathbb{R} \times [0,\infty)$. Define

$h_2(x_1,x_2) = (x_1,x_2)$ for $x_2 \le x_1$,

$h_2(x_1,x_2) = (-x_2 + 2x_1,x_1)$ for $x_1 \le x_2$.

$h_2$ maps the line segment connecting $(0,x_2)$ and $(x_2,x_2)$ bijectively onto the line segment connecting $(-x_2,0)$ and $(x_2,x_2)$. It is easy to verify that $h_2$ is continuous. Next define $g: \mathbb{R} \times [0,\infty) \to [0,\infty)^2$ by

$g(y_1,y_2) = (y_1,y_2)$ for $y_2 \le y_1$,

$g(y_1,y_2) = (y_2,2y_2 - y_1)$ for $y_1 \le y_2$.

$g$ is continuous and $g \circ h_2 = id$, $h_2 \circ g = id$. Therefore $h_2$ is a homeomorphism.

For $n \ge 2$ we therefore obtain a homeomorphism

$$h_n = h_2 \times id_{[0,\infty)^{n-2}} : [0,\infty)^n \to \mathbb{R} \times [0,\infty)^{n-1} .$$

Then

$$h = (id_{\mathbb{R}^{n-2}} \times h_2) \circ ... \circ (id_{\mathbb{R}} \times h_{n-1}) \circ h_n$$

is the desired homeomorphism.

Answer 2

I tried to prove using induction as follows: For case $n=1$, $[0,\infty) = \Bbb{H}^1$, and for $n=2$ the map $f(z)=z^2$ will do the job, regard $\Bbb{R}^2$ as the complex plane $\Bbb{C}$. Now assume that $\mathbb{H}^k\approx \bar{\mathbb{R}}^k_+$ for all $2< k \leq n$. We have \begin{align} \bar{\mathbb{R}}^{n+1}_+ = \bar{\mathbb{R}}^{n}_+ \times \bar{\mathbb{R}}_+ &\approx \mathbb{H}^n \times \bar{\mathbb{R}}_+ = \mathbb{R}^{n-1} \times \bar{\mathbb{R}}_+ \times \bar{\mathbb{R}}_+ \\ &\approx \mathbb{R}^{n-1} \times \mathbb{H}^2 = \mathbb{R}^{n-1} \times \mathbb{R} \times \bar{\mathbb{R}}_+ = \mathbb{R}^n \times \bar{\mathbb{R}}_+ = \mathbb{H}^{n+1}. \end{align}

In fact, this proof can be extend to arbitrary topological space. Suppose that $X$ and $Y$ be any topological spaces. Let $H^n:=X^{n-1}\times Y$. If $H^2 \approx Y^2$, then by induction \begin{align} H^{n+1} &= X^{n-1}\times X \times Y = X^{n-1}\times H^2 \\ &\approx X^{n-1}\times Y^2 = H^n \times Y \\ &\approx Y^n\times Y \\ &=Y^{n+1} \end{align}