While reading expository material on homotopy theory, I've come across the concept of globular sets, and I was wondering to what extent these can, informally speaking, "take on the role of simplicial sets" for algebraic topology?

Specifically: there is already the well-known notion of the singular simplicial set functor $$ \mathscr{S} : \mathrm{Top} \rightarrow \mathrm{sSet}\;, $$ described (on e.g. nLab) as the functor $X \mapsto (\,[n] \mapsto \mathrm{Hom}_{\mathrm{Top}}(\delta([n])\,,X)\,)$. Here $\Delta$ is the simplex category, $\mathrm{sSet} = \mathrm{Set}^{\Delta^{\textrm{op}}}$, and $\delta : \Delta \rightarrow \mathrm{Top}$ is the "standard cosimplicial object in $\mathrm{Top}$", given (as I understand it) by sending the face and degeneracy maps to the corresponding maps between the standard geometric simplices, in the expected way.

Is it possible to set up some sort of a "singular globular set functor" $$ \mathscr{G} : \mathrm{Top} \rightarrow \mathrm{gSet}\;, $$ given by the functor $X \mapsto (\,[n] \mapsto \mathrm{Hom}_{\mathrm{Top}}(\theta([n])\,,X)\,)$? Here $\mathbb{G}$ is the globe category (using the definition at e.g. https://ncatlab.org/nlab/show/globe+category), $\mathrm{gSet} = \mathrm{Set}^{\mathbb{G}^{\textrm{op}}}$, and $\theta : \mathbb{G} \rightarrow \mathrm{Top}$ is some "standard coglobular object in $\mathrm{Top}$".

(E.g., could we define the functor $\theta : \mathbb{G} \rightarrow \mathrm{Top}$ as sending $[n]$ to the $n$-disk, and sending $[n] \xrightarrow{\sigma,\tau}[n+1]$ to the maps $D^n\rightarrow D^{n+1}$ which are inclusions as the upper and lower boundary hemispheres respectively?)

If such a "singular globular set functor" $\mathscr{G}$ does make sense, does it admit a left adjoint "geometric realization functor", similarly to the singular simplicial set functor $\mathscr{S}$?

If not, is there some notion of a "geometric realization functor" $\mathrm{gSet}\rightarrow\mathrm{Top}$, even if it is not left adjoint to some singular globular set functor $\mathscr{G}$?

(Apologies in advance if these questions are elementary or not well-formed; I haven't studied homotopy theory in great depth.)

  • 2
    $\begingroup$ Yes, you can define such functors, in that way. The nerve–realisation adjunction is very general. The real question to ask is whether these functors are useful. In the context of homotopy theory, one could ask e.g. is there a transferred model structure on the category of globular sets making the adjunction a Quillen equivalence? (I think the answer is no.) $\endgroup$
    – Zhen Lin
    Apr 21, 2021 at 8:14

1 Answer 1


As Zhen Lin indicates, there is no problem defining such functors, but unfortunately they do not allow us to "do homotopy theory" purely in the world of globular sets.

The technical version of this statement is that the globe category is not a "test category" in the sense of Grothendieck (the citation on that nlab page is to Scholium 8.4.14 in Cisinski's thesis which is in French). This means that there is no known way to define weak equivalences of globular sets in such a way that the homotopy category is the same as the homotopy category of spaces. There are also theorems telling us that this implies there is definitely no way to do this and at the same time attain certain further nice properties, but I am a bit hazy on the precise statements. See Cisinski's thesis, or section 4 of Ara for an overview in English.

Roughly, the reason is that the globe category is not "rich" or "complicated" enough to encode all the homotopical information one would like. Intuitively, one should expect that in order to model homotopy theory, your category should encode composition of paths and higher composition as well. For instance, in the simplex category, there is a morphism $\Delta[1] \to \Delta[2]$ which "misses" the middle vertex; this is the morphism encoding composition of paths in simplicial sets. The globe category has no such morphisms. Rather, when globular sets are used to model higher categories, the composition needs to be added as extra structure over and above the globular set itself. The nlab article I linked to suggests (without citation) that if such morphisms are added by passing to the free monoidal category on the globe category, one does obtain a test category.

  • $\begingroup$ Thank you for the very informative answer! $\endgroup$ Jun 2, 2021 at 22:57
  • 4
    $\begingroup$ Scholium 8.4.14 actually gives a proof that taking the free monoidal category on the globe category (with the 0-globe as unit) gives a test category. The reason why globular sets cannot model homotopy types is that fundamental groups of globular sets are always free groups. $\endgroup$ Jul 11, 2022 at 19:29

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