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Let $A$ be an $n$-dimensional alternative real division algebra (not necessarily associative). Is the map $$ \eta \colon \bigl\{(x,y) \in A \times A : |x|^2+|y|^2=1\bigr\} \to A \sqcup \{\infty\}, \quad (x,y) \mapsto \frac{x}{y} $$ with $x/0 :=\infty$ continous? Let me explain a few things:

  • Division Algebra: For every $a \neq 0$ and $b$ there exists $x,y$ with $b = ax$ and $b=ya$. You should think about $x$ and $y$ as some kind of fractions $\frac{b}{a}$, but wait... there are two? Thats why we have the second property
  • Alternative: $a(ab) = a^2b$ and $(ab)b = ab^2$ for every $a$ and $b$.
  • One can easily see that alternatively implies that every element $\neq 0$ has a unique multiplicative two-sided inverse, so we get a well defined map $\eta$.
  • We give $A \sqcup \{\infty\}$ the one-point-compatification topology, that means its open sets are the open subsets of $A$ and sets $G \sqcup \{\infty\}$ where $G \subseteq A$ has compact complement in $A$.
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