Covering space action
I am having problems with 1.3.24 in Hatcher's Algebraic Topology:
24. Given a covering space action of a group G on a path-connected, locally path-connected space X, then each subgroup H in G determines a composition of covering spaces X -> X/H -> G. Show:
a. Every path-connected covering space between X and X/G is isomorphic to X/H for some subgroup H in G
The best I can do here is say that since we have a covering space action then pi_1(X/H1) = pi_1(X/H2); not sure how to proceed.
b. Two such covering spaces X/H1 and H/H2 of X/G are isomorphic iff H1 and H2 are conjugate subgroups of G.
c. The covering space X/H -> X/G is normal iff H is a normal subgroup of G, in which case the group of deck transformations of this cover is G/H.
(definition of covering space actionat (*) on page 72 in this pdf http://www.math.cornell.edu/~hatcher/AT/ATch1.pdf)
For (a), Since X is a path-connected, locally path-connected space and the action of G is a covering space action, is a subgroup of . By using a Galois's correspondence, there exists a covering space X/H of X/G such that is a subgroup of . In a similar vein, X is the covering space of X/H.
Originally Posted by harbottle
For (b), an isomorphism between covering spaces and is a homeomorphism such that . This implies that for each , f preserves the covering space structure, taking to .
If H_1 and H_2 are conjugate subgroups of G, then so are and in , where and . The remaining step is to show that there exists a bijection between each point and for each point .
For (c), if is normal, then is a normal subgroup of . The group of deck transformation is N(L)/L (Hatcher p71, proposition 1.39b). Since L is a normal subgroup of , we have . It remains to show that . This equation intuitively makes sense, but I can't think of the rigorous proof for now. Can you proceed from here?
Thank you very much for this. Some of the concepts are still new to me so I need to take a little while to work out what everything means; I'll work through your solution and see what I can do.
Originally Posted by aliceinwonderland