Timeline for The classifying space of an infinite totally ordered set is contractible

6 events

when toggle format	what		by	license	comment
May 17, 2015 at 13:54	comment	added	Todd Trimble		So if (nonempty) $P$ merely has binary joins, the same argument as in the second paragraph shows $BP$ is contractible. This hadn't occurred to me before now.
May 17, 2015 at 13:01	comment	added	Neil Strickland		The two maps combine to give a poset map $\{0,1\}\times P\to Q$, and $B(\{0,1\}\times P)=[0,1]\times BP$.
May 17, 2015 at 12:34	comment	added	Tatjana Popow		Thank you. Do you have a reference for the standard fact about homotopic maps on classifying spaces induced by order-preserving maps between posets?
May 17, 2015 at 12:16	comment	added	Neil Strickland		The space $BP$ is topologised as the colimit of the spaces $BQ$, as $Q$ runs over finite subsets of $P$. So a subset $F\subset BP$ is closed iff $F\cap BQ$ is closed with respect to the obvious topology on $BQ$, for all $Q$.
May 17, 2015 at 12:11	comment	added	Tatjana Popow		Which topology should one use on the set of maps $x$, which are identified with $BP$? The one coming from the (possible infinite) product $[0,1]^P$?
May 17, 2015 at 11:53	history	answered	Neil Strickland	CC BY-SA 3.0