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Dec 27, 2017 at 7:47 answer added Taras Banakh timeline score: 3
Dec 22, 2017 at 22:03 answer added Prof. David Edwards timeline score: 0
Dec 22, 2017 at 21:52 comment added fosco This is a good answer too; what is known about the shape theory of this space?
Dec 22, 2017 at 21:48 comment added André Henriques Asking for the weak homotopy type of $\mathbb{QP}^\infty$ is the wrong question to ask, because that topological space does not have enough maps $\mathbb R^n\to \mathbb{QP}^\infty$. Instead, one should ask for its shape: en.wikipedia.org/wiki/Shape_theory_(mathematics)
Dec 22, 2017 at 21:22 vote accept fosco
Dec 22, 2017 at 21:12 answer added Eric Wofsey timeline score: 19
Dec 22, 2017 at 21:10 answer added Tim Campion timeline score: 10
Dec 22, 2017 at 20:44 comment added fosco (although there are a few things I would like to see in detail: why is this lifting possible -i.e. is the projection map a covering like in real and complex case?-)
Dec 22, 2017 at 20:43 comment added fosco @FanZheng maybe you saying that a path $\gamma : I \to \mathbb{QP}^\infty$ lifts along the projection from $\mathbb Q^\infty$? And the domain now is totally disconnected, I agree.
Dec 22, 2017 at 20:37 comment added fosco @FanZheng I'm not sure about anything, I'm just sitting here waiting for Eric to answer :-)
Dec 22, 2017 at 20:35 comment added Fan Zheng A path from $(a,\dots)$ to $(b,\dots)$ projects to a path from $a$ to $b$ in $\mathbb Q$, which is impossible if $a\neq b$.
Dec 22, 2017 at 20:33 comment added Tim Campion I would guess that it's totally path-disconnected, and so homotopically discrete. But that's just a guess.
Dec 22, 2017 at 20:22 history asked fosco CC BY-SA 3.0