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Apr 13, 2021 at 22:52 answer added Nicholas Kuhn timeline score: 3
Apr 13, 2021 at 19:52 answer added Neil Strickland timeline score: 5
Apr 13, 2021 at 19:46 answer added user171227 timeline score: 5
Apr 12, 2021 at 13:42 comment added Ann Oh, thanks! It's an interesting example. So, $j_*$ is not injective in general. But $H \mathbb Q$ is not of finite type. May be, it's true for such spectrum $E$?
Apr 12, 2021 at 3:31 comment added Eric Peterson @user171227 I don't think that's cheating!
Apr 12, 2021 at 0:19 comment added user171227 Maybe this is cheating, but $[H\mathbb{Q},\Sigma H\mathbb{Z}]= \mathrm{Ext}(\mathbb{Q},\mathbb{Z})$ is torsion free and non-zero.
Apr 12, 2021 at 0:06 history asked Ann CC BY-SA 4.0