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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!

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

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

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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 $H=p^{-1}(\mathbb{D}_n)$.

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    $\begingroup$ 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. $\endgroup$
    – Aloizio Macedo
    Oct 15, 2016 at 18:52
  • $\begingroup$ @AloizioMacedo, you're right, I've just edited it. Thanks. $\endgroup$
    – rmdmc89
    Oct 15, 2016 at 19:01

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