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Stream: theory: category theory

Topic: an Ex-citing diversion

Amar Hadzihasanovic (Apr 08 2020 at 14:19):

Morgan Rogers said:

I said "simplicial complex", intending "geometric realisation of the category", since Quillen's theorem also talks about the classifying space. I don't know homotopy theory well enough to know how a fibrant replacement of a simplicial set is constructed.

Anyway, there's a “quick and kind of useless” way which is to take the singular complex of the geometric realisation, and a “more complicated but very useful” combinatorial way involving an iteration of Kan's $\mathrm{Ex}$ functor, the adjoint of barycentric subdivision.

Amar Hadzihasanovic (Apr 08 2020 at 14:21):

I think that what I said should be provable using the latter construction, but this is the point where I would rather just look up the proof :D

Paolo Capriotti (Apr 08 2020 at 14:23):

Right, $\mathrm{Ex}$ is useful here. It suffices to show that $\mathrm{Ex}(NC)$ is contractible. An $n$-sphere in $\mathrm{Ex}(NC)$ is the same as a map $\mathrm{sd}(\partial \Delta[n]) \to NC$, and a cocone over this diagram is the same as a map $\mathrm{sd}(\Delta[n]) \to NC$, i.e. a filler for the original sphere in $\mathrm{Ex}(NC)$.

Morgan Rogers (he/him) (Apr 08 2020 at 14:24):

I think we should keep spaces in the mix for the purposes of computing examples, because I reckon that I should get a homotopy-equivalent space from any suitable presentation of my monoid (with a 1-simplex for each generator and suitable cells to provide homotopies between related composites), which for finitely generated monoids would be a massive advantage.

Morgan Rogers (he/him) (Apr 08 2020 at 14:25):

Paolo Capriotti said:

Right, $\mathrm{Ex}$ is useful here. It suffices to show that $\mathrm{Ex}(NC)$ is contractible. An $n$-sphere in $\mathrm{Ex}(NC)$ is the same as a map $\mathrm{sd}(\partial \Delta[n]) \to NC$, and a cocone over this diagram is the same as a map $\mathrm{sd}(\Delta[n]) \to NC$, i.e. a filler for the original sphere in $\mathrm{Ex}(NC)$.

I will need all of the notation explaining, if you don't mind :sweat_smile:

Morgan Rogers (he/him) (Apr 08 2020 at 14:25):

ie how are you constructing Ex and/or how can I think about it?

Amar Hadzihasanovic (Apr 08 2020 at 14:25):

Paolo Capriotti said:

Right, $\mathrm{Ex}$ is useful here. It suffices to show that $\mathrm{Ex}(NC)$ is contractible. An $n$-sphere in $\mathrm{Ex}(NC)$ is the same as a map $\mathrm{sd}(\partial \Delta[n]) \to NC$, and a cocone over this diagram is the same as a map $\mathrm{sd}(\Delta[n]) \to NC$, i.e. a filler for the original sphere in $\mathrm{Ex}(NC)$.

Thanks! Is there a reason why doing a single round of $\mathrm{Ex}$ is sufficient?

Amar Hadzihasanovic (Apr 08 2020 at 14:27):

Morgan Rogers said:

ie how are you constructing Ex and/or how can I think about it?

By adjointness: $n$-simplices in $\mathrm{Ex}(X)$ are the same as maps $\mathrm{sd}(\Delta[n]) \to X$, i.e. barycentrically subdivided simplices in $X$...

Amar Hadzihasanovic (Apr 08 2020 at 14:29):

There is a way of including $X$ into $\mathrm{Ex}(X)$ via a natural way of mapping the subdivision of a simplex onto the simplex (all the maximal simplices but one get tossed onto a vertex)

Morgan Rogers (he/him) (Apr 08 2020 at 14:30):

Where sd means subdivision and $\Delta[n]$ is your notation for the n-simplex, okay.
You say "natural", but I can see at least $n$ ways of doing that :rolling_on_the_floor_laughing:

Amar Hadzihasanovic (Apr 08 2020 at 14:30):

I think there's exactly two of them that are functorial :)

Paolo Capriotti (Apr 08 2020 at 14:30):

Amar Hadzihasanovic said:

Thanks! Is there a reason why doing a single round of $\mathrm{Ex}$ is sufficient?

I was thinking something along these lines: you prove that if $X$ is a filtered simplicial set (i.e. satisfies a right orthogonality property analogous to the definition of filtered for categories), then $\mathrm{Ex}(X)$ is contractible, and this is basically the argument above. Now by induction all the $\mathrm{Ex}^n(NC)$ are contractible, so the colimit should also be contractible.

Morgan Rogers (he/him) (Apr 08 2020 at 14:31):

What is the "right orthogonality property" you're referring to?

Reid Barton (Apr 08 2020 at 14:32):

These facts about Ex can be found in Kerodon 3.3.5 and its containing section

Paolo Capriotti (Apr 08 2020 at 14:32):

something like right lifting property with respect to maps $N(J) \to N(J) \ast \Delta[0]$, where $J$ is a finite category

Paolo Capriotti (Apr 08 2020 at 14:33):

where $\ast$ is join

Morgan Rogers (he/him) (Apr 08 2020 at 14:37):

... join? Do you mean $N(J) * \Delta[0]$ is the cone under/over $N(J)$ in the geometric sense?

Paolo Capriotti (Apr 08 2020 at 14:37):

yes

Morgan Rogers (he/him) (Apr 08 2020 at 14:38):

I've never seen that given as "the" definition of filtered before, but then much of the notation/terminology used in this topic has been a little alien to me. I can see how this is a convenient form for this definition in this context.

Paolo Capriotti (Apr 08 2020 at 14:40):

But since $X \to \mathrm{Ex}(X)$ is a weak equivalence (as that Kerodon page shows), you actually don't need the whole inductive argument above: if $\mathrm{Ex}(X)$ is contractible, then (the fibrant replacement of) $X$ is contractible.

Amar Hadzihasanovic (Apr 08 2020 at 14:42):

Maybe I'm wrong, but I think a gap with the argument above is that you can't check contractibility by only “filling boundaries of standard n-simplices” unless you are in a Kan complex

Amar Hadzihasanovic (Apr 08 2020 at 14:42):

Which is why you need to look at the $\partial \Delta[n] \to \mathrm{Ex}^\infty(NC)$

Amar Hadzihasanovic (Apr 08 2020 at 14:43):

But these are going to land into $\mathrm{Ex}^k(NC)$ for a finite $k$, so then I think you can just apply your argument with $\mathrm{sd}^k(\partial \Delta[n])$

Paolo Capriotti (Apr 08 2020 at 14:43):

Filling boundaries automatically gives that it is Kan complex, because the map $X \to 1$ is a trivial fibration (hence a fibration).

Amar Hadzihasanovic (Apr 08 2020 at 14:46):

Ah of course! That's the step I was missing. :)

Amar Hadzihasanovic (Apr 08 2020 at 14:48):

Is it always the case that $\mathrm{Ex}(NC)$ is a Kan complex, or is there something special about $C$ being contractible?

Paolo Capriotti (Apr 08 2020 at 14:55):

I don't think it is always the case. It is probably related to some Ore condition or similar. But I'm not sure.

Paolo Capriotti (Apr 08 2020 at 14:56):

For example, if $C= \bullet \leftarrow \bullet \to \bullet$, then it seems you cannot fill 2-horns in $\mathrm{Ex}(NC)$. But double check :)

Morgan Rogers (he/him) (Apr 08 2020 at 14:58):

Should I ask what a 2-horn is..?

Paolo Capriotti (Apr 08 2020 at 15:00):

A 2-dimensional horn

Paolo Capriotti (Apr 08 2020 at 15:00):

i.e. two lines glued by a vertex

Morgan Rogers (he/him) (Apr 08 2020 at 15:03):

That sounds 1-dimensional! But okay, I'm relieved that it's simpler than I expected.

Paolo Capriotti (Apr 08 2020 at 15:08):

Yes, it's really 1-dimensional, but the index is 2, because you always think of the $n$-horns $\Lambda^i[n]$ it as the domains of the horn inclusions $\Lambda^i[n] \to \Delta[n]$, and $\Delta[n]$ is $n$-dimensional, of course. Here $i$ is the face that you removed from the boundary. Being a Kan complex means that every map $\Lambda^i[n] \to X$ can be extended to $\Delta[n]$.

sarahzrf (Apr 08 2020 at 15:09):

hmm, seems inconsistent with the fact that, say, the n-sphere is the boundary of the n+1-ball...

sarahzrf (Apr 08 2020 at 15:09):

:)

Mike Shulman (Apr 08 2020 at 19:23):

But it's consistent notationally with the simplicial $n$-sphere, namely $\partial \Delta[n+1]$.