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1:11 AM
@TimCampion could this be what you want?
https://www.maths.ed.ac.uk/~v1ranick/papers/doldpup2.pdf
 
2:01 AM
Why is a milnor square (i.e. we have ring maps row 1: A->B, row 2: A'->B', columns are surjection, and the square is a pull back in abelain groups) a pullback square of E_1 ring spectra?

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I am following the notation in p1 (for milnor square ) of the following paper by Land and Tamme, https://arxiv.org/pdf/1808.05559.pdf
The claim that this is a pb square in Alg_E_1(sp) is line 2 of page 2.
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Originally my argument goes like this: it is a pullback square in spectra iff it is a push out. Thus we compute \pi_i of the tensor product (A \otimes_A' B) .
 
 
2 hours later…
3:43 AM
OK, here is an argument I believe is true:
In the above setting, let I=Ker(A->A') (1-categorically).
Then
1. We have fiber sequence I\otimes_A B -> A \otimes_A B ->A' \otmes_A B
2. All these are connective spectra.
3. At \pi_0 we get what we want.
4. \pi_i for the middle term vanishes for all i>0, hence we deduce that A' \otimes_A B is in fact a discrete spectra isomoprhic to B'.

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I am happy for any comments/ if this argument is incorrect
 
 
4 hours later…
7:19 AM
@BryanShih Why are you using ⊗? The pullback in E_1-ring spectra is computed in spectra
You just need to check that the vertical fibers are equivalent, which is automatic from the π_0 surjectivity and the fact that it is a pullback of abelian groups
 
7:32 AM
@DenisNardin That's a much simpler argument! My proof goes as:
1. the square is pushout in spectra
2. pushout=pullback
Is this bogus?
 
The outline seems sound, but I don't understand why you're using ⊗ here?
It feels like you are trying to compute the pushout in E_∞-rings, but that's of course not what you're supposed to do (not least of all because a priori none of the objects in the square is an E_∞-ring :D)
 
8:09 AM
Ah yes you are compltely right - That was a terrible mistake! I thought the tensor product of these E_1 rings gives the pushout in spectra. Thanks a lot. Now just completely forget my previous arguments.
 
 
1 hour later…
9:36 AM
@BryanShih I think that's exactly it. Thanks!
 

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