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In some categories of things with interesting structure, said structure can be recovered from the category.

For example, in the category of chain complexes of abelian groups, if you're given a chain complex $A$, you can recover its structure as a chain complex from all of the hom-groups $\hom(C_n, A)$, where the $C_n$ are all of the shifts of the complex $\mathbf{Z} \to \mathbf{Z}$.

By the yoneda lemma, the same goes for simplicial sets; you can recover the simplicial set structure of $S$ from the hom-sets $\hom(\mathbf{y}\Delta^n, S)$ and the maps between them.

I've been curious for some time if this sort of thing would help me get a handle on $\infty$-categories.

If I try the simplicial set recipe on $\infty \mathrm{Grpd}$, though, I get a problem; if I've not made an error, all of the basic simplicies are homotopic, and all of the maps homotopy equivalences! e.g. an output of this recipe could well be the constant simplicial object.

It's reassuring, I suppose, that I get a simplicial object that obviously corresponds to the object I started with, but doesn't actually help with what I was trying to do!

I suppose the essential difference between this and the previous examples is here I'm (attempting to) work in an $\infty$ category, but the previous examples were all ordinary categories. But that leaves me at a loss; I want to see how we could use the $\infty$ category structure to produce things like simplicial sets from the objects, but I'm at a loss at how to proceed.

Thus, the question: can you recover a simplicial set from an $\infty$ groupoid? And how?

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  • $\begingroup$ Exactly what do you mean by "recovering" the simplicial set? If you want a procedure that produces a simplicial set given an $\infty$-groupoid that will likely depend from the particular model of $\infty$-groupoids you are using. For example, if you are using Kan complexes as a model, taking the underlying simplicial set will work perfectly. To be clear: the question you are asking could be rephrased as: can we get a simplicial set from an homotopy type? The answer is yes but the precise procedure will depend on the model you are using for homotopy types. $\endgroup$
    – Denis Nardin
    Commented Jul 26, 2016 at 14:46
  • $\begingroup$ Could you explicit the recipe on infinity groupoids? And what model of infinity groupoids are you using (Kan fibrant? globular?)? $\endgroup$
    – user40276
    Commented Jul 26, 2016 at 14:47
  • $\begingroup$ @DenisNardin: Yes; you can rephrase my question as how to go from a homotopy type to a simplicial set. I was hoping there was something you could do that was model-independent; at the very least an overall recipe you could follow with any model even if you needed a model to follow the actual steps of the recipe. $\endgroup$
    – user13113
    Commented Jul 26, 2016 at 14:58
  • $\begingroup$ @user40276: I'm intentionally trying not to pick a model. The calculation that leads to the constant simplicial set would be that if I had some collection of $\infty$ groupoids $\Delta^n$ I consider as 'being' the standard simplicies, all of the $\hom(\Delta^n, G)$ would be equivalent to $\hom(\Delta^0, G)$, so if I have any idea at all of how things work, the simplicial object I get from the $\hom(\Delta^n, G)$ should be equivalent to the on I get by instead using $\hom(\Delta^0, G)$ in degree $n$. $\endgroup$
    – user13113
    Commented Jul 26, 2016 at 15:05
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    $\begingroup$ @Hurkyl I think that what you want is the theory of minimal Kan complexes. Every $\infty$-groupoid has a unique minimal Kan complex that models it. I'm not sure how explicit is the construction though $\endgroup$
    – Denis Nardin
    Commented Jul 26, 2016 at 15:11

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You can think of simplicial sets as presentations of $\infty$-groupoids; for example, in any simplicial set modeling $BG$ for $G$ a discrete group, say with one 0-simplex, the 1-simplices give generators of $G$ and the 2-simplices give relations between them. An object can have many presentations in general, and so it's unclear how to pick between them.

You might object that in the case of $BG$ there is in some sense a canonical presentation, namely the monadic one: every element of $G$ as a generator, every relation between them as a relation, etc. If you try to run the same game on an arbitrary pointed connected $\infty$-groupoid $X$ what you will write down is a simplicial space (not set), more or less the usual bar construction but using $\Omega X$ in place of $G$.

Anyway, continuing the theme of your examples, the model-independent question would've been to ask how to recover an $\infty$-groupoid from an object in the $\infty$-category of $\infty$-groupoids, and the answer is very simple: it's the entire hom $\infty$-groupoid $\text{Hom}(1, X)$ (note that whenever you talked about a hom you implicitly only talked about its $\pi_0$).

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  • $\begingroup$ N.B. it's the entire hom ∞-groupoid Hom(1,X) as Qiaochu Yuan says and that is exactly the proper generalisation of the method you mention. $\endgroup$
    – Tim Porter
    Commented Jul 26, 2016 at 20:00
  • $\begingroup$ I see, I hadn't thought of it as constructing a presentation. So to construct something presenting $G$, I could pick a set to be the degree 0 things and map it into G. Taking the pullback should give me the groupoid of paths between my chosen points. Iterating should give me a globular set, and I just need to take care that I pick up all of the components at each step. I imagine I ought to be able to work out a variation to get a simplicial or cubical set.... $\endgroup$
    – user13113
    Commented Jul 29, 2016 at 5:52
  • $\begingroup$ I know that we should have $X^1 \cong X$, but that's sort of the opposite of what my goal was, since I wanted to see how one might arrive at the external structure of the sorts of objects we use as a model through purely internal concerns. I suppose I didn't convey that aim as well as I had hoped in my question. $\endgroup$
    – user13113
    Commented Jul 29, 2016 at 5:52
  • $\begingroup$ @Hurkyl: no, you communicated that aim, but my claim is that this is not a model-independent question. If you want to find a simplicial set presenting a given $\infty$-groupoid my claim is that you need to make some choices somewhere. $\endgroup$
    – Qiaochu Yuan
    Commented Jul 29, 2016 at 5:55
  • $\begingroup$ @Hurkyl: I don't understand your construction. Can you write it up in more detail? $\endgroup$
    – Qiaochu Yuan
    Commented Jul 29, 2016 at 6:02

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