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Tim Campion
Tim Campion
10:13 AM
nice
 
Tim Campion
Tim Campion
10:34 AM
By the way, I don't think Lior would mind me pointing out that the MO question I linked to (here it is again) was his -- I think some of these considerations may have made it into some of his work on ambidexterity or at least some stuff he's done with $\pi$-finite spaces.
(In general when somebody uses a pseudonym on MO, it's safest to assume that they really don't want to be publicly identified, but in this case Lior publicly identified himself to me on MO at one point.)
(That was a funny story too. He asked a question, and I left a comment saying "Tomer and his students have been thinking about this -- you should ask them")
 
 
8 hours later…
jdc
jdc
7:06 PM
@PeterHaine This is a very nice write-up, and seems very natural and optimal
I had a question about the first display in Construction 2.5, which I couldn't follow completely. It seemed to me that if n is 3 and X_1, X_2, and X_3 are connected, then the left-hand side should have eight components but the right-hand side only five. Do you perhaps mean something more subtle by the product in Spc_* than the product in Spc, pointed at the product of the basepoints?
 
 
3 hours later…
Peter Haine
Peter Haine
9:43 PM
@jdc Thanks!
Thanks for pointing this out. You're definitely right. I hadn't checked this carefully since it was only to do the counterexample in model-independent way, but I totally messed up the combinatorics! I think the right expression is that
X₁₊ × ⋯ Xₙ₊ ≃ ∨_{∅ ≠ I ⊂ {1,…,n}} (∏_{i ∈ I} Xᵢ)₊ .
So if you split off the I = {1,…,n} and #I = 1 terms in this expression you get
X₁₊ × ⋯ Xₙ₊ ≃ (∏_{i ∈ I} Xᵢ)₊ ∨ (∐_{i ∈ I} Xᵢ)₊ ∨ [∨_{I ⊂ {1,…,n}, 1 <#I <n} (∏_{i ∈ I} Xᵢ)₊ ].
(So the two terms I wrote, plus a bunch of other junk.)
 

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