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Oct 11, 2022 at 22:11 comment added Neil Strickland I think that the proof implicitly gives a specific zigzag of equivalences joining $Q$ to any other reasonable model of the same homotopy type. Given such a zigzag, there is a contractible space of choices of how to straighten it.
Oct 11, 2022 at 22:03 comment added André Henriques I guess, in your language, my question becomes whether $Q$ is defined in such a way that there is a canonical (up to contractible choice) projection map $Q\to K(\mathbb Z,2n)$, for every $n$...
Oct 11, 2022 at 20:19 history answered Neil Strickland CC BY-SA 4.0