# Open hemisphere is connected

It is intuitively clear that an open hemisphere of $\mathbb{S}^n$, say $H=\{(x_1, ..., x_{n+1})\in \mathbb{S}^{n}\mid x_{n+1}>0\}$, is connected. However, when I'd tried to formalize it, I've realized I couldn't really do it.

My first idea was to prove that $H$ is path-connected, which implies connectedness. That is also intuitively clear, but it turned out to be much more complicated to write down than I expected.

Any suggestions? Thanks!

Consider two points $x,y$ in the hemisphere, and take the straight line $p$ connecting both in $\mathbb{R}^{n+1}$.

What is $\widehat{p}=\frac{1}{\Vert p\Vert }p$?

You can think about a sort of projection of $H$: Let $\mathbb{D}_n=\{(x_1, ...,x_n)\in \mathbb{R}^n \mid x_1^2+...x_n^2<1\}$ and define $p:H\to \mathbb{D}_n$ with $p(x_1, ...,x_{n+1})=(x_1, ...,x_n)$. It's easy to see that $p$ is continuous bijection with continuous inverse given by $(x_1, ...,x_n)\mapsto (x_1, ...,x_n, \sqrt{1-x_1^2-...-x_n^2})$. Since $\mathbb{D}_n$ is open and connected, so is $H=p^{-1}(\mathbb{D}_n)$.

• There is a small detail. Being a "continuous bijection" is not enough to conclude that the inverse image is connected. For that, you need that the inverse function is continuous in your argument. Oct 15, 2016 at 18:52
• @AloizioMacedo, you're right, I've just edited it. Thanks. Oct 15, 2016 at 19:01