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Jan 18 at 10:48 vote accept Aidan
Jan 18 at 2:16 history edited Chris Gerig CC BY-SA 4.0
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Jan 18 at 2:15 comment added Chris Gerig You're right, updated response.
Jan 18 at 2:11 history edited Chris Gerig CC BY-SA 4.0
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Jan 18 at 2:04 history edited Chris Gerig CC BY-SA 4.0
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Jan 18 at 1:59 history edited Chris Gerig CC BY-SA 4.0
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Jan 18 at 1:19 history edited LSpice CC BY-SA 4.0
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Jan 18 at 0:55 comment added Andy Putman What does the Kan-Thurston theorem have to do with this? We already know that $X$ is a $K(G,1)$.
Jan 18 at 0:55 comment added Dmitri Pavlov What exactly is the relevance of the Kan–Thurston theorem here? The OP explicitly states that X is a classifying space for a discrete group G, which by definition makes it weakly equivalent (and not merely homology equivalent) to K(G,1).
Jan 18 at 0:50 history answered Chris Gerig CC BY-SA 4.0