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S Nov 29, 2017 at 20:31 history suggested Mike Pierce CC BY-SA 3.0
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Nov 29, 2017 at 19:29 review Suggested edits
S Nov 29, 2017 at 20:31
Nov 16, 2009 at 13:02 comment added Jon Woolf The problem if U is not connected is that Z_V(U) need no longer vanish; if V is a component of U then Z_V(U) is a summand of Z(U). So this proof doesn't work without connectivity (at least not without modification).
Nov 16, 2009 at 2:38 comment added Andrew Critch Can't the above proof be rewriten with the word "connected" removed?
Nov 13, 2009 at 18:03 comment added David Treumann In the first sentence of your proof you use assume that X is locally connected--is that necessary? I was just about to edit my answer to hedge a little more and say that I don't know what's going on with the Cantor set.
Nov 13, 2009 at 17:51 history answered Jon Woolf CC BY-SA 2.5