5
$\begingroup$

Let $\mathcal{H}$ be a Grothendieck $(\infty,1)$-topos. According to this page in nlab, for any $X \in \mathcal{H}$, the suspension object $\Sigma X$ is homotopy equivalent to the smash product $B \mathbb{Z} \wedge X$, where $B \mathbb{Z}$ is the "classifying space of the discrete group of integers." Furthermore, for any pointed object $X \in \mathcal{H}_*$ and any group object $G \in Grp(\mathcal{H})$, the article says we can "form the tensor product $X \otimes G \in Grp(\mathcal{H})$."

My problem is: none of this terminology is explained, nor does the page provide a reference. Specifically, what is $\mathbb{Z}$ in an arbitrary $\infty$-topos? What is the smash product $\wedge$? What is the tensor product $\otimes$? My best guess is that $\otimes$ refers to the unique tensor structure on $\mathcal{H}_*$ such that the map $\mathcal{H} \to \mathcal{H}_*$ is symmetric monoidal (here $\mathcal{H}$ is given the Cartesian monoidal structure), but this is only a guess.

Is there a reference where all these notions are defined?

$\endgroup$
  • 6
    $\begingroup$ I think that $\mathbb{Z}$ in an $(\infty,1)$-topos is the constant sheaf associated to the ordinary $\mathbb{Z}$. $\endgroup$ – David Roberts Feb 28 at 2:02
  • 1
    $\begingroup$ The object $X$ in your first paragraph should be pointed, I think. Then as $\mathbf{B}\mathbb{Z}$ is also pointed, I would imagine that the smash product is defined in the usual way as a homotopy quotient of the product by the wedge sum, which is itself a homotopy pushout. $\endgroup$ – David Roberts Feb 28 at 2:18
  • 4
    $\begingroup$ Every $(\infty,1)$-topos has an nno, by applying the inverse image part of the canonical geometric morphism to Spaces, and inverse image functors preserve the relevant structure to build the integers from the natural numbers. Which is pretty much what you said, but in a model-independent way :-). Or else, form the circle as a homotopy pushout of the 'usual diagram' of finite sets' and take its loop space. This will give the integers, too, much like in HoTT, but now the object is a bit fatter, and not a strict group object. $\endgroup$ – David Roberts Feb 28 at 4:56
  • 3
    $\begingroup$ @DavidRoberts You can totally say model-independently what a strict group object is: it is an algebra for the Lawvere theory of (strict) groups (if you want, the homotopy category of the Lawvere theory of "nonstrict" groups). It's fairly straightforward to see that left-exact functors preserve algebras for Lawvere theories and so $\mathbb{Z}$ is a strict group object in every topos. $\endgroup$ – Denis Nardin Feb 28 at 8:50
  • 1
    $\begingroup$ @DenisNardin: In my understanding, by “strict groups” people usually really mean strict with respect to the on-the-nose equality of some model, which isn’t and can’t be a model-invariant notion. What you’re describing is what I’d see as the model-invariant/homotopy-invariant replacement of the concept — which I agree is the better thing to think of, in general, but I don’t think it’s quite what people (at least people I know/read) would usually mean by “strict groups”. $\endgroup$ – Peter LeFanu Lumsdaine Feb 28 at 12:30
5
$\begingroup$

To summarize the comments so far:

  • In an arbitrary $(\infty,1)$-topos $\mathcal{E}$, the integers can be defined as the loop space of the circle $S^1_\mathcal{E}$, which itself is given as the (homotopy) pushout of two copies of the map $\mathrm{pt}\sqcup \mathrm{pt} \to \mathrm{pt}$ in $\mathcal{E}$. Otherwise, one can calculate them as the image of the usual $\mathbb{Z}$ under the inverse image $\mathcal{S}paces \to \mathcal{E}$ of the canonical map to $\mathcal{S}paces$. Note that in this definition, $\mathbf{B}\mathbb{Z}$ is $S^1_\mathcal{E}$, and has a canonical (up to equivalence) basepoint.

  • The smash product of a pointed object $X$ in $\mathcal{E}$ and $S^1_\mathcal{E}$ is defined as the (homotopy) coequaliser of the two canonical maps $$ X \vee S^1_\mathcal{E} \mathrel{\mathop{\rightrightarrows}^{\mathrm{pt}}_{\mathrm{incl.}}} X\times S^1_\mathcal{E} $$ where the wedge sum $X \vee S^1_\mathcal{E}$ is the (homotopy) pushout of $X\leftarrow \mathrm{pt} \to S^1_\mathcal{E}$.

$\endgroup$
  • $\begingroup$ I do have one more question. Is it easy to see that a left adjoint $F: \mathcal{H}_* \to Grp(\mathcal{H})$ to the functor $ \Omega B: Grp(\mathcal{H}) \to \mathcal{H}_*$ should always exist? One would have to show at least that $\Omega B$ preserves limits, but if $Grp(\mathcal{H})$ is not presentable I don't see why this would be enough $\endgroup$ – leibnewtz Feb 28 at 8:06
  • 1
    $\begingroup$ @leibnewtz Well, $Grp(\mathcal{C})$ is always presentable if $\mathcal{C}$ is presentable (since it is a Bousfield localization of the presentable category $\mathrm{Fun}(\mathrm{Fin}_*,\mathcal{C})$) $\endgroup$ – Denis Nardin Feb 28 at 8:28
  • $\begingroup$ Isn't it easier to just say that $\mathbb{Z}_{\mathcal{E}}$ is the coproduct of $\mathbb{Z}$-many copies of the terminal object? I would expect $\Omega S^1 \simeq \mathbb{Z}$ to be a theorem, not a definition... $\endgroup$ – Mike Shulman Mar 1 at 0:27
  • $\begingroup$ @Mike sure, that occurred to me, too, but if one is being picky about minimal machinery to construct the integers, using finite (co)limits might be cleaner. But then, I suppose, one does want to know how $\Omega S^1_\mathcal{E}$ relates to $\Delta \Omega S^1 \simeq \Delta \mathbb{Z}$, but surely they are equivalent by the exactness and cocontinuity of $\Delta$, and the construction of $\Omega S^1$? $\endgroup$ – David Roberts Mar 1 at 4:40
  • $\begingroup$ @DavidRoberts Yeah, since $\Delta$ is left exact you just need to verify it for the topos of spaces. There it is a theorem, ofc, that $S^1$ is 1-truncated with $\pi_1$ given by $\mathbb{Z}$. I don't know if there's any proof using only HoTT machinery. $\endgroup$ – Denis Nardin Mar 1 at 9:29

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.