@AaronMazel-Gee @RuneHaugseng do I need some kind of condition on simplicial colored operads to take the operadic nerve and get the right thing? some kind of "fibrancy" condition?

I feel like one of you guys said this recently in here...

For some reason I feel it's crucial, when passing from monoidal model categories to monoidal quasicategories that in the mapping spaces $Mul(X_1\otimes\ldots \otimes X_n,Y)$ that the source was cofibrant and the target was fibrant...

But I'm just misremembering something.

say you're in a simplicial model category. if you want the hom-ssets to be presenting the correct hom-spaces, then yes you need the source to be cofibrant and the target to be fibrant. if you're mapping out of a tensor product as $\mathrm{hom}(X_1 \otimes \cdots \otimes X_n,Y)$ -- which corresponds to the multimorphism object $\mathrm{Mult}(\{X_1,\ldots,X_n\},Y)$ -- then of course you'll want $X_1 \otimes \cdots \otimes X_n$ to be cofibrant and $Y$ to be fibrant.

for the former, it suffices for all the $X_i$ to be cofibrant (and that's probably really the only way you'll ensure this, in practice) -- this follows from the axioms for a monoidal model category

Right. Okay, that's what I thought. Okay.

Thanks!

Is there a general way to interpret what people mean when they say they want something to be "homotopically well behaved"?

I guess... homotopy invariant?

"homotopy invariant" means, maybe, that something is a functor on some homotopy category. but you can and might want to ask for more than this, which is that it's a functor on some infinity-category

but i guess someone is saying to me that something needs to be homotopically well behaved so that it passes to the relevant infinity category.

so i sort of am wondering what that means before the infinity category comes into play

Or, actually I guess the phrase is "maintain homotopical control."

dunno. some kind of fibrancy / cofibrancy thing?

maybe you should just quote what the person is saying

Just checking, I think this is true, but is the multiplication map XxX-->X of a grouplike simplicial monoid a fibration?

Couldn't you take something like the nerve of a weakly contractible endofunctor category to get a counterexample?

@JonBeardsley View the natural numbers as a category with one object and the natural numbers as endomorphisms. The nerve of this category is a simplicial monoid, but the multiplication map in not a fibration. (exercise: find an explicit (outer) 2-horn lifting problem that can't be solved).

Oh and it is a connected simplicial set and hence grouplike.

Aha. Thanks!

This would be solved by having a homotopy inverse?

For some reason I was thinking grouplike induced a homotopy inverse, but looking at that example, it seems unlikely to have any kind of "inverse" map

Like, very roughly, I have this intuitive sense that the homotopy inverse should make is so that all the fibers are homotopy equivalent.

maybe this thing isn't a kan complex?

i don't actually know very much simplicial stuff

Ah interesting point.

Yeah... the thing I'm thinking of is definitely a Kan complex.

(being imprecise again sorry)

Yeah, if X isn't a Kan complex, even I can think of counterexamples.

Is there any categorical significance to the usual formal definition of functions (in Set) $f:A \to B$ as subsets of AxB such that the left projection is a bijection? I'm imagining that in an arbitrary category with finite products one could desire a map $Hom(A,B) \into S$ where S is the set of all subobjects $i:D \into AxB$ for which $pi_1 \circ i$ is an isomorphism

@barron when you write $AxB$ do you mean $A\times B$? with \times instead of x?

yes. I don't have chat tex enabled so none of it renders for me anyway, I get sloppy

(fwiw I don't have anything helpful to say about your question, unfortunately)

Oh ok, no worries

As far as I know such definitions are usually historical, i.e. they describe a bunch of examples that people want to describe in some generality, so the litmus test for me is whether or not the thing you're describing has interesting examples attached to it.

I'll try to work it out in Top and Mon

it is true in any category with finite products, though (of course) one has to identify isomorphic subobjects in order to get an actual bijection

an amusing corollary is that a well-powered category with finite products must be locally small

What's the argument (to show any category with finite products satisfies this)? I'm not seeing it.

there are evident maps in both directions – just show that they are mutually inverse

darn, it took me a couple of weeks of slow background thinking to understand that this question of mine mathoverflow.net/questions/228016/… from a couple of weeks back is so confused. I would edit it very much, but I fear that would render Peter May's answer obsolete. Dunno what I should do. Already pinged the relevant people (Peter May and Denis)

@ZhenLin is this resolved by requiring X to be a Kan complex?

I have no idea

Ah, fair enough.

@lentic do you know Mandell-May-Schwede-Shipley? there, enriched functors on CW-complexes are called "W-spaces" if I remember correctly and they're compared with various other models

I had only vaguely looked at it, I'm looking at the relevant stuff in there now, thanks!

is it equivalent for a functor to be linear (pointed 1-excisive) to it carrying cofibration sequences to fibration sequences? (i.e. only converting some special homotopy pushouts to homotopy pullbacks)

is this a functor from pointed spaces to pointed spaces?

yes

from finite CW, if you want

it seems reasonable to me. I mean, it's a space-version of the deduction of Mayer-Vietoris from excision

the stuff surrounding the unstable adams spectral sequence is rly complicated, i'm having a very hard time wrapping my head around some of it