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Greetings everyone,

I'm a little confused about what open sets in $SO(3,\mathbb{R})$ essentially "look like".I'm trying to show that $f : SO(3,\mathbb{R}) \rightarrow S^{2}$ given by $A \mapsto Ae_1$ is a fibre bundle which is homeom. to the circle $S^{1}$.

Now I know that all matrices in $SO(3,\mathbb{R})$ have orthonormal columns so the map $f$ is surjective and continuous, but if I have an open set $U \in S^{2}$ (which is easy to visualize) then what does $f^{-1}(U) \in SO(3,\mathbb{R})$ look like?

I think visualizing this in some way will help me show that it is homeom. to $U \times S^{1}$.

Any ideas?

Thanks!

Post Closed as "too localized" by Deane Yang, S. Carnahan
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# What do open sets in $SO(3,\mathbb{R})$ look like?

Greetings everyone,

I'm a little confused about what open sets in $SO(3,\mathbb{R})$ essentially "look like". I'm trying to show that $f : SO(3,\mathbb{R}) \rightarrow S^{2}$ given by $A \mapsto Ae_1$ is a fibre bundle which is homeom. to the circle $S^{1}$.

Now I know that all matrices in $SO(3,\mathbb{R})$ have orthonormal columns so the map $f$ is surjective and continuous, but if I have an open set $U \in S^{2}$ (which is easy to visualize) then what does $f^{-1}(U) \in SO(3,\mathbb{R})$ look like?

I think visualizing this in some way will help me show that it is homeom. to $U \times S^{1}$.

Any ideas?

Thanks!