I found Seveneves a bit hard to read... the first two parts were rather bleak, after all

Yeah they were seriously bleak...

I did some significant tearing up while reading those parts.

Given a 1-simplex in the quasicategory of Kan complexes, how much indeterminacy is there in picking a morphism in in category of simplicial sets with the Quillen model structure which recovers that 1-simplex after applying the simplicial nerve? I can always choose a Kan fibration right?

@AaronMazel-Gee Hello, this is sort of a random question. I saw on the PSI page that you spent some time there. How was it? I love the ocean and the place looks awesome!

@JonBeardsley so if the quasicategory of Kan complexes is the simplicial nerve of the fibrant simplicial category of Kan complexes, then in a precise sense there's no indeterminacy at all, but that's an evil thing to say because of course if you chose an equivalent edge you might get a different-but-equivalent map between different-but-equivalent simplicial sets

in most cases, if you're considering a morphism in isolation rather than as part of some bigger diagram, you're free to choose any map of simplicial sets in the equivalence class

in particular, you're free to choose a fibration

of course, you really do have to allow yourself the possibility of changing one of the endpoints

Anyone know if there's anything in HA about taking an O-monoidal quasicategory and looking at the O-monoidal structure on an overcategory? Specifically, do you get it for free? What do tensor products look like?

That's discussed in section 2.2.2 - you need an O-algebra structure on the object you're slicing over

Ah right, that makes sense. Thanks!

@RuneHaugseng maybe I'm just not seeing it, but is the forgetful map from the overcategory over an O-algebra O-monoidal? I don't seem to see anything about that in this section.

We have these two coCartesian fibrations $C^\otimes_{/X}\to O^\otimes$ and $C^\otimes\to O^\otimes$, and we have a map of quasicategories $C^\otimes_{/X}\to C^\otimes$. Maybe it's obvious that this gives a commuting triangle.

2.2.4(2) tells you it's at least lax monoidal

Probably if you unwrap the description of the coCartesian morphisms in Lemma 2.2.2.8 you'll see the forgetful functor preserves all of them...

Ah, ok so a functor that preserves inert morphisms is lax O-monoidal? And if it preserves inert AND coCartesian morphisms it's strongly O-monoidal?

Umm, isn't that the definition?

Well, I was guessing it was, but I can't find a definition of "lax O-monoidal functor" in HA, haha.

It's just a morphism of infinity-operads over O

You'd think there'd be an index entry like... "lax monoidal" or something.

Oh ok. Sure. I'll have to collate those two concepts in my brain, lol.

So just one that preserves inert morphisms (i.e. coCartesian morphisms over inert ones)

Almost as effective as being wrong about things in a talk.

maybe a stupid question: is the monoidal structure induced by join of simplicial sets symmetric? It seems pretty strange since it doesn't restrict to symmetric on Cat

@tetrapharmakon It's not. Consider $(\Delta^0\coprod\Delta^0)*\Delta^0$ vs $\Delta^0*(\Delta^0\coprod\Delta^0)$, which are non-isomorphic simplicial sets.

yes (nlab warns that ordinal sum is not symmetric monoidal)

can you spot where something nonsymmetric happens in the formula $(X\star Y)_n = \coprod_{i+j=n}X_i\times Y_j$?

The non-symmetry appears in the structure maps.

ah, there's a dedicated MO thread

like always :-)

thx

can I deduce that $X\star Y$ and $Y\star X$ have the same sequence of sets but different faces and degeneracies inducing on them a splcl structure?

("the same" up to iso, say)

Their sets of n-simplices are in bijection for each n, but they have different structure maps, yes.

nice, now I understand :-)

Did you see Tyler's answer here? mathoverflow.net/questions/7134/joins-of-simplicial-sets

yup, I was precisely working in that direction, Tyler avoided me to overthink

it's the evilest trick in simplicial stuff!

Suppose I have a map of spaces $X\to Y$, where $X$ is contractible and both are $n$-fold loop spaces. Then is this map necessarily (equivalent to) a map of $n$-fold loop spaces?

@Jon: yes. you can replace X with a point. a point has a unique n-fold loop space structure, and the unique map pt -> Y sending pt to the basepoint of Y is an n-fold loop map

so the answer is iff X lands in the identity component of Y