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I know that there is a short exact sequence (where all maps are continuous): \begin{align} \mathbb{Z}/2\mathbb{Z} \rightarrow \text{Spin}(n) \rightarrow \text{SO}(n) \end{align}

Showing that $f: \text{Spin}(n)/\{1,-1\} \rightarrow \text{SO}(n)$ is a group isomorphism. This group isomorphism is also continuous by the universal property of the quotient topology, but why is $f$ a homeomorphism? Maybe I'm overlooking something obvious here but I'd be glad to get some help.

Henry T.
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    $\text{Spin}(n) \to SO(n)$ is a two-fold covering map, so you can use that. Alternately, any continuous bijection from a compact space onto a Hausdorff space is a homeomorphism. – David Gao May 04 '24 at 02:41
  • The problem is that I need this result to show that it is a two-fold covering. Do you know how I could show that Spin is compact without using that it is a cover for SO(n)? – Henry T. May 04 '24 at 02:42
  • I mean, that depends on how you define $\text{Spin}(n)$. Some people define it as the two-fold covering of $SO(n)$. If you define it using Clifford algebra, though, I believe it’s not too hard to show it is a closed and bounded subset of the Clifford algebra, which is finite-dimensional, so the result follows from that. – David Gao May 04 '24 at 02:53
  • Thank you! Then I'll try to understand how I'd define a norm on the Clifford algebra under which Spin then is bounded. – Henry T. May 04 '24 at 03:18
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    You could apply a more general result such as this: https://math.stackexchange.com/questions/3760064 – Lukas Heger May 04 '24 at 03:40

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