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Prove $S^2= \{x \in \mathbb{R}^{3}: \lVert x \rVert =1 \}$ is not a topological group, I know that in this answer and lie group can solve it, but I need a form to use a result in differential topology: $f: S^2 \rightarrow S^2$ continuous exists $x \in S^2$ such that $f(x)=x$ or $f(x)=-x$.

So, if $S^2$ is a topological group then $g:S^2 \rightarrow S^2$, $g(x)=x^{-1}$ is a homeomorphim function, but I don't know how use the result.

PSW
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Since $S^2$ is compact, any function that is continuous is uniformly continuous. Assume that $S^2$ is a topological group, take $\omega \in S^2$ near enough to the identity such that $|x\cdot\omega -x\cdot e|<\frac{1}{2}$ for every $x$ (here I am taking the usual euclidean metric and using that multiplication is uniformly continuous), since $\omega \neq e$, the function $f(x) = \omega \cdot x$ does not satisfy $f(x) = x$ for any $x$, hence, there is $x \in S^2$ such that $f(x) =-x$, but notice that $$ \frac{1}{2}>|f(x)-x\cdot e| = |-x-x| = 2|x| = 1 $$ which is a contradiction.

Raul Fernandes Horta
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  • This seems to use a result that every continuous map $f$ from $S^2$ to itself either has point $x$ such that $f(x) = x$ or a point such that $f(x) = -x$ (or both). What is this result? The Borsuk-Ulam theorem? – Vincent May 27 '24 at 11:16
  • @Vincent yes, is Borsuk-Ulam – Raul Fernandes Horta May 27 '24 at 18:44
  • Right, I remember it as talking about functions from $S^2$ to $\mathbb{R}^2$. I have to think a bit about how to get from there to this version about functions from $S^2$ to itself – Vincent May 28 '24 at 16:04