Homotopy Theory

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Harry Gindi

12:00 AM

so you just call them I think algebraic morphisms between logoi

mathieu.anel.free.fr/mat/doc/Anel-Joyal-Topo-logie.pdf

Dylan Wilson

12:13 AM

@AaronMazel-Gee I really appreciate you sharing this, as well as your raw description of the emotions and sensations that arose for you when the event occurred. I completely agree that we should celebrate one another much more than we do, instead of this 'competitive' point of view or 'zero sum' attitude. What would math look like if we were trained in sympathetic joy? (in buddhism it's called 'mudita'; you can read more about it here: dharmanet.org/coursesM/16/bv3.htm)

4 hours later…

Alexander Campbell

4:04 AM

@AaronMazel-Gee Hear, hear.

@HarryGindi Yes this is true. One way to prove this is to use that the classes (bijective on objects, strictly fully faithful) form an ordinary factorisation system on the 1-category of quasi-categories.

I say that a morphism of simplicial sets $f \colon A \to B$ is "strictly fully faithful" if the induced map $A \to B \times_{cosk_0 B} \mathrm cosk_0 A$ is an isomorphism of simplicial sets; i.e. if for every $n \geq 1$, every $(n+1)$-tuple of objects $a_0,\ldots,a_n$ of $A$, and every $n$-simplex $\beta$ of $B$ with vertices $f(a_0),\ldots,f(a_n)$, there exists a unique $n$-simplex $\alpha$ of $A$ with vertices $a_0,\ldots,a_n$ such that $f(\alpha) = \beta$.

You also use that any e.s.o. functor of quasi-categories is equivalent to a monic b.o. functor, and any fully faithful functor of quasi-categories is equivalent to a strictly fully faithful isofibration.

3 hours later…

Harry Gindi

7:08 AM

Thanks!!!

Harry Gindi

7:29 AM

Heya

I just wanted everyone to know who also cares about mathematics. I've been banned from MO for a full year with no warning.

Rune Haugseng

7:41 AM

You also have such a factorization system on the infinity-category of Segal spaces: the functor Seg(S) -> S taking a Segal space X to its space X_0 of objects is a cartesian fibration (via pullback along the right adjoint, given by RKE) and a fully faithful morphism of Segal spaces is exactly a cartesian morphism for this fibration. So you get a factorization system from the factorization of a morphism as morphism in a fibre + cartesian.

(Note this does not descend to complete Segal spaces/infinity-categories, but I think it induces the factorization as essentially surjective + fully faithful there.)

Denis Nardin

@AaronMazel-Gee I am terribly sorry to have caused you this. It really was not my intention -- for what is worth I wasn't trying to establish priority, just to give a quick reference and this was the one I'm most familiar with, for various reasons. In the interest of not doing it again, can you tell me your paper where you prove B-K, because I don't think I know which one it is...

Rune Haugseng

This paper by Edoardo Lanari discusses such factorization system from cartesian fibrations, by the way: arxiv.org/abs/1911.11533

Harry Gindi

Thanks!

7 hours later…

Dedalus

2:32 PM

I am confused. Say I have a noetherian ring R and an ideal I generated by a regular sequence (a_1,...,a_n). I have a natural map of inverse systems {R/(a_1^m,..., a_n^m)} -> {R/I^m}. Why is this a pro-isomorphism in this situation?

Dedalus

2:48 PM

Nevermind

Harry Gindi

@Dedalus what was the answer?

Do you just test it by evaluating on an arbitrary representable pro-ring??

Dedalus

3:50 PM

@HarryGindi I am not sure I am correct anymore since I do not seem to use the hypothesis that the ideal I is regular, so
either the authors don't need the hypothesis or I screwed up.
I use Proposition 2.3 of https://core.ac.uk/download/pdf/82323187.pdf . Call the map R/(a_1^m,..,a_n^m) -> R/I^m for
f_m and the map of inverse systems for f. Then according to the proposition, f is an isomorphism iff for each m,
I can find a s > m and a map g:R/I^s -> R/(a_1^m,...,a_n^m) satisfying the requirements of the proposition.

Harry Gindi

4:02 PM

thanks

1 hour later…