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If one wanted to find all connected covering spaces of a product of two spaces, say $S^1\times RP(3)$, how would you go about it?

I'm thinking finding the fundamental group of $S^1\times RP(3)$, and then we know that to each subgroup of $\pi_1(S^1\times RP(3)$(there is s 1-1 correspondence between coverings of a space $X$ and conjugacy classes of subgroups of $\pi_1(X)$) corresponds a covering space of $S^1\times RP(3)$. Is this correct?

If yes, then how would I go about identifying the corresponding covering spaces? Would I simply try to identify which spaces have these subgroups as their fundamental groups? If I found all these subgroups, would this imply that there cannot be any more connected covering spaces because of this one-to-one correspondence?


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Just a caution, since it's happened before around this time, that it's finals season for grad students (not that I'm assuming anything about abby, I'm just saying...) – Dylan Wilson Nov 29 '12 at 17:13
I'm not a graduate student, and it's not the finals season either...but I understand your concern. I'm not looking for an answer, just to see whether I'm on the right track. – abby Nov 29 '12 at 19:06
Okay- well, start by finding the universal cover, then use the correspondence between quotients of that by nice group actions and covers of the original space. – Dylan Wilson Nov 30 '12 at 4:17
(That is the explicit form of the correspondence you mention between conjugacy classes of subgroups of the fundamental group and covers.) – Dylan Wilson Nov 30 '12 at 4:18

1 Answer 1

alternative approach given covering space actions of $G_1$ on $X_1$ and $G_2$ on $X_2$ the action of $G_1\times G_2$ on $X_1 \times X_2$ defined by $(g_1,g_2)(x_1,x_2)=(g_1(x_1),g_2(x_2))$ is a covering space action and $X_1\times X_2/{G_1\times G_2}$ is homeomorphic to $X_1/G_1 \times X_2/G_2$.

so for this problem if we take $G_1 = \mathbb{Z} , G_2= \mathbb{Z_2}, X_1= \mathbb{R}, X_2= S^3$ then we are done.

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