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1:02 AM
Hey @CharlesRezk do you know of any reference in which a Quillen equivalence between pointed connected simplicial sheaves on a site, and grouplike monoids in simplicial sheaves, is proven?
I know of a reference for this fact for \infty-topoi in Higher Algebra, but not for the model category case, and it seems like it is probably written down somewhere.
 
 
2 hours later…
2:44 AM
@JonathanBeardsley I think Goerss-Jardine, sections V.6 and V.7 are related to your question in the case of the trivial site. (I don't know any references for more general sites, but that doesn't mean much.)
 
 
7 hours later…
9:35 AM
Silly question: is there a name for functors F : C ---> D with the following property? For every pair of objects x and y in C, if there is an isomorphism F(x) ---> F(y) in D, there is an isomorphism x ---> y in C.
This is not quite a functor that "reflects isomorphisms", since I am not requiring the first map to be of the form F(f).
 
@PedroTamaroff To clarify: are you also not asking that the iso x --> y be sent by F to the given iso Fx --> Fy?
 
Not really, no.
Just that some iso exists.
Perhaps one can arrange it so that this happens, but that's not relevant for me.
 
Then the best I can come up with is "injective on $\tau_0$", since the set of isomorphism classes of objects of a category $C$ is sometimes denoted by $\tau_0(C)$.
 
I see.
I am leaning towards calling it "rigid", but not sure if this term is taken
 
There are "rigid monoidal categories", but I don't know about "rigid functors"...
 
 
3 hours later…
12:21 PM
Does anyone know a reference for the fact that $Cat^x$ (the ∞-category of stable ∞-categories and exact functors) has direct sums?
(I know how to prove it, I just don't want to write down a proof of this standard fact in the paper)
 
 
2 hours later…
2:28 PM
@DenisNardin I might look in Blumberg-Gepner-Tabuada...
@PedroTamaroff I'd be tempted to say "essentially injective", since it's dual to the well-established "essentially surjective". But ultimately I'd probably prefer Alexander's suggestion.
 
3:00 PM
I'm looking at Bullett-Macdonald paper where they prove a reformulation of the Adem relations (sciencedirect.com/science/article/pii/…) and I got terribly confused by something. At the start of the second page they say that if you take the product $y_1 \cdots y_n x_{n+1} \cdots x_{2n}$ where $y$'s are the exterior generators of $H^{\ast}(B\mathbb{Z}/p)$ and $x$'s are the polynomial generators.
Then this element induces an injection of $H^{\ast}(K(\mathbb{Z}/p,3n)) \to H^{\ast}(B (\mathbb{Z}/p \times \cdots \times \mathbb{Z}/p))$ in degrees $\le 4n$. I haven't checked this but it certainly seems likely. Then they go on to claim that because of this it suffices to prove any relations in the mod p steenrod algebra on this element $y_1 \cdots y_n x_{n+1} \cdots x_{2n}$.
I'm finding it really hard to see why, Is it some combinatorial thing I'm missing here? Naively you would want some injection in ranges like $\le O(p)$ to make sure you get all the Adem relations introduced by a new generator. What am I missing here?
 
 
5 hours later…
8:05 PM
Why $O(p)$?
 
I was being sketchy. What I meant is that when you go up in degree in the steenrod algebra you see a new generator every $2(p-1)$ degrees. So my initial naive guess is that you'll be safe if you have agreement on cohomology in the ranges up until the first new generator appears. But now that I've thought about this a bit this is a very crude heuristic. The relations that a new generator of degree $2n(p-1)$ introduces are always in degrees $\ge 2n(p-1)+2(p-1) = 2(n+1)(p-1)$.
But $2(n+1)(p-1)/2n(p-1) = (n+1)/n \le 4/3$ (for $n\ge3$) which makes everything seem fine now.
 
8:26 PM
Actually, I'm wrong again. It's still not obvious that this can capture all the relations. The largest adem relation introduced by a new generator happens in degree $2n(p-1) + 2\floor{\frac{n}{p}}(p-1)$. So what I really want is $2n(p-1) / 2\floor{\frac{n}{p}}(p-1) = \frac{n}{\floor{\frac{n}{p}}} \le \frac{n}{\frac{n}{p} +1}$ to be $\le 4/3$ for sufficiently large $n$ which doesn't seem to be correct.
If everytime a generator in degree $2n(p-1)$ pops you get a relation in degree $2n(p-1) +2\frac{n}{p}(p-1)$ how can you detect it if you can only detect cohomology operations that can jump $4/3$ times the degree you are start in.
 
Just make sure the degree you start in is much bigger than n?
 
I'm confused. What is your $n$?
Maybe the claim is that $\frac{4}{3} * (2n(p-1) - 1)$ (i.e. start right before the generator appears) is $\ge \frac{n}{\frac{n}{p}+1}$
That can work
I think it does...
Sorry I meant $\frac{4}{3}(2n(p-1) -1) > 2(n-1)(p-1)+2\floor{\frac{n-1}{p}}(p-1) $
Actually I think this doesn't work either.
I give up :(
Nevermind, I think its actually okay. Sorry for the spamming.
Funnily enough it seems like the $4/3$ ratio is sort of optimal in this case.
I guess the question now is how can you know in advance that $4/3$ will work if you don't know the adem relations. Or do they just mean that once you know where the adem relations are you can a posteriori check them using this.
 
9:00 PM
I don't really understand what is the issue. It's easier to talk about in the prime 2 case, where $H^*K(Z/2, n)\to H^*B(Z/2)^n$ is injective for $*\leq 2n$. So if you have two additive stable cohomology operations $a,b\colon H^*\to H^{*+k}$, to determine whether they are equal apply them to the class $x_1\cdots x_n\in H^nB(Z/2)^n$, as long as you choose $n>k$.
 
9:11 PM
Why is it important that it's injective for $\leq 2n$ then? Could it be $\leq n$ just as well?
I was sure that $2$ has to do with the fact that you want to apply the square to the universal class and still land inside the injective range.
If the cohomology operation applied to the universal class sends it outside of the injective range than how can you know anything about it using this?
That's why my initial guess for what you would want to have in char p is a map which is injective in a range like $\le 2np$ so that you could apply the reduced power and still be inside the injective range. But then I realized it's more subtle than that... I'm still extremely open to the possibility I'm missing something completely obvious here
 
9:32 PM
These are stable operations, so you can choose n to be whatever you want.
 
Yeah. I just realized that.
The char 2 spoiled me.
For p=2 you can get away without using the stability. But in fact it's enough to have a map to $K(\mathbb{Z}/p,n)$ injective in a range $\le cn$ as long as $c >1$.
Then you can use stability to push everything to $n \gg 0$ until you can detect the operation. The 4/3 doesn't really matter.
Took me long enough. Thanks for the help!
 

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