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

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.

• I think the problem is the line $$f(\overline{\mathbb{R}}^n_+ \setminus \{ (0,\dots,0)\}) =f( (0,\infty) \times \dots\times (0,\infty))$$ as the two sets whose f-images you are taking are NOT equal. For example, $(1,0,\dots,0) \in \overline{\mathbb{R}}^n_+ \setminus \{ (0,\dots,0)\}$ but $(1,0,\dots,0) \notin (0,\infty) \times \dots\times (0,\infty)$ – itinerantleopard Jul 18 '18 at 3:12
• To see that they are indeed homeomorphic, it might help to think about some small-dimensional cases first. When $n=1$, the space $\overline{\mathbb{R}}_+^1$ is actually equal to $\mathbb{H}^1$. When $n=2$, the space $\overline{\mathbb{R}}^2_+$ is the closed upper right quadrant of the plane and $\mathbb{H}^2$ is the closed upper half plane. The former can be homeomorphically deformed into the latter by "unfolding" the boundary until it lies flat against the $x$-axis. The geometric picture in higher dimensions is similar. Hope that helps! – itinerantleopard Jul 18 '18 at 3:18

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.

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}

• The proof of $\bar{\mathbb{R}}_+ \times \bar{\mathbb{R}}_+ \approx \mathbb{H}^2$ ($n = 2$) is missing. It cannot be reduced to $n = 1$. – Paul Frost Jul 19 '18 at 12:15
• @PaulFrost You're right. I didn't see that before. Thank you. I'll amend my answer. – Sou Jul 19 '18 at 12:24
• I like your proof for $n = 2$. (+1) – Paul Frost Jul 19 '18 at 17:38
• @PaulFrost : Thanks. I like your proof too. – Sou Jul 19 '18 at 17:40