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Apr 27, 2015 at 21:41 comment added Allen Knutson You're not even using $G$ a group, just the map $\pi_1(X/H) \to H$ from the usual theory of covering spaces (here $X \to X/H$).
Apr 27, 2015 at 13:58 comment added Neil Strickland You can continue the fibre sequence as $H\xrightarrow{i} G\xrightarrow{j} X\xrightarrow{k} BH\xrightarrow{l} BG$, and then $\delta=\pm k_*$.
Apr 27, 2015 at 12:20 history asked Mikhail Borovoi CC BY-SA 3.0