My question concerns the proof of Proposition 4.2 in Bhatt-Mathew’s paper on the arc-topology, but my confusion is completely general and anyone familiar with limits in $\infty$-categories would know what to do. The situation is that they have a map of $R$-algebras $V\to \tilde V:=V_{\mathfrak p}\times V/\mathfrak p$ which is a covering map for their Grothendieck topology, and they consider an $\infty$-category $\mathcal C$ which is compactly generated by cotruncated objects and a functor $\mathcal F:R\mathrm{-alg}\to\mathcal C$ which preserves finite products ($V$ is a valuation ring with prime ideal $\mathfrak p$ and residue field $\kappa(\mathfrak p)$, but this is not important). They then claim that the totalisation $\lim\mathcal F(\tilde V^{\otimes\bullet+1})$ coincides with the pullback $\mathcal F(V_{\mathfrak p})\times_{\mathcal F(\kappa(\mathfrak p))}\mathcal F(V/\mathfrak p)$. Here, $\tilde V^{\otimes\bullet+1}$ means the cosimplicial $R$-algebra $\tilde V\rightrightarrows\tilde V\otimes_V\tilde V\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow}\ldots$. It’s important that in this situation $V_{\mathfrak p}\otimes_V V_{\mathfrak p}=V_{\mathfrak p}$, $V/\mathfrak p\times_VV/\mathfrak p=V/\mathfrak p$, $V_{\mathfrak p}\otimes_VV/\mathfrak p=\kappa(\mathfrak p)$, and $\kappa(\mathfrak p)\otimes_VV_{\mathfrak p}=\kappa(\mathfrak p)=\kappa(\mathfrak p)\otimes_VV/\mathfrak p$. In fact, once you make these identifications, you can forget what $V$ and $\mathfrak p$ are and use the fact $\mathcal F$ preserves products to write down $\mathcal F(\tilde V^{\otimes\bullet+1})$ very concretely; the first few terms are $$\mathcal F(V_{\mathfrak p})\times \mathcal F(V/\mathfrak p)\rightrightarrows\mathcal F(V_{\mathfrak p})\times \mathcal F(V_{\mathfrak p}\otimes_V V/\mathfrak p)\times\mathcal F(V/\mathfrak p\otimes_VV_{\mathfrak p})\times\mathcal F(V/\mathfrak p)\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow}\ldots$$ where both of the terms with a tensor product are canonically equivalent to $\mathcal F(\kappa(\mathfrak p))$ and all the maps are the canonical ones.

I’m having trouble seeing the claim that this limit coincides with the pullback. (If it were a diagram in a 1-category it would be completely obvious.) The claim that $\mathcal C$ is compactly generated by cotruncated objects means that it suffices to show the result after applying $\operatorname{Map}_{\mathcal C}(x,-)$ and in this way replace $\mathcal C$ by $\mathcal S_{\leq n}$, the full subcategory of $\mathcal S$ on $n$-truncated spaces. Then we can truncate the above diagram so it is indexed by $\Delta_{\leq n+2}$ and deal with a finite diagram. The two approaches I’ve tried are to compute the above (infinite) totalization as a homotopy limit in $\mathcal S$ and to break up the products and show some finite sub-diagram of the resulting diagram is coinitial (left cofinal) in this specific situation. More specifically, I’ve tried to show the diagram $D$ is coinitial, where $D$ has four objects $\mathcal F(V_{\mathfrak p})$, $\mathcal F(V/\mathfrak p)$, and two copies of $\mathcal F(\kappa(\mathfrak p))$ and four non-identity arrows: one arrow each from the first two objects to each copy of the second. However, this doesn’t appear to be true. On the other hand, I haven’t found a nice enough way to express the homotopy limit in $\mathcal S$ to allow me to make the desired identification. Any tips would be a big help, thank you.

Edit A simplified model: objects $X,Y,Z$ with maps $X\to Y$, $Z\to Y$ in an $\infty$-category with limits and a tensor product $\otimes$ which commutes with $\times$; we have $X\otimes Y=Y=Z\otimes Y$, $X\otimes Z=Y$, $X\otimes X=X$, $Y\otimes Y=Y$, $Z\otimes Z=Z$, and we consider the cosimplicial object $(X\times Z)^{\otimes\bullet+1}$ with first three terms $$X\times Z\rightrightarrows X\times (X\otimes Z)\times (Z\otimes X)\times Z\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow}X\times (X\otimes X\otimes Z)\times(X\otimes Z\otimes X)\times(Z\otimes X\otimes X)\times (Z\otimes Z\otimes X)\times(Z\otimes X\otimes Z)\times(X\otimes Z\otimes Z)\times Z,$$ i.e. $$X\times Z\rightrightarrows X\times Y\times Y\times Z\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow}X\times Y^{\times 6}\times Z.$$ Breaking apart the product, the three maps above all send $X\to X$, and then send $X=X\otimes X\to X\otimes X\otimes Z$, $X\otimes Z\otimes X$, or $Z\otimes X\otimes X$.

The claim is that the limit of the above diagram is $X\times_YZ$.

  • 3
    $\begingroup$ I haven't read the background you wrote in detail, but does the following resolve your question? If $X \to Y \leftarrow Z$ is a span of pointed objects in an $\infty$-category with limits, then $X\times_Y Z$ is the totalization of the cosimplicial object $X \times Y^\bullet \times Z$. $\endgroup$
    – skd
    Aug 6, 2022 at 21:04
  • $\begingroup$ Keyword(s) : Bousfield-Kan $\endgroup$ Aug 7, 2022 at 8:23
  • $\begingroup$ @skd I added a simplified model. I think it’s a little different from $X\times Y^\bullet\times Z$ because that has $2^n$ copies of $Y$ in the product in the $n$th degree, whereas here we have 2, then 6, and in general $2^{n+1}-2$. $\endgroup$
    – Tomo
    Aug 9, 2022 at 19:29
  • $\begingroup$ @MaximeRamzi In terms of Bousfield-Kan I’m familiar with the comparison formula expressing the homotopy limit of the above diagram $X^\bullet$ of Kan complexes (which is Reedy fibrant) as $\operatorname{Fun}(\Delta^\bullet,X^\bullet)$, where $\operatorname{Fun}$ is the internal hom in $\mathcal C^\Delta$; i.e. the simplicial set with $\operatorname{Fun}(\Delta^\bullet,X^\bullet)_n=\operatorname{Hom}(\Delta^\bullet\times\Delta^n,X^\bullet)$, here the product is levelwise and the hom is in $\mathrm{SSet}^\Delta$. I don’t see how to see this computes the fiber product in $\mathcal C$. $\endgroup$
    – Tomo
    Aug 9, 2022 at 19:34
  • $\begingroup$ Tomo : Bousfield-Kan to express a pullback in cosimplicial terms :) $\endgroup$ Aug 9, 2022 at 19:41

1 Answer 1


Let $X$, $Y$, and $Z$ be Kan complexes. We wish to show that $X\times_YZ$ in $\mathrm{Spc}$ can be computed as the limit of the diagram $$(*)\qquad X\times Z\rightrightarrows X\times Y\times Y\times Z\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow}X\times Y^{\times 6}\times Z\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow \\[-1em] \rightarrow}\ldots$$ in $\mathrm{Spc}$. This will follow from claim that in $\mathrm{Fun}(\Lambda_2^2,\mathrm{Spc})$, the geometric realization (i.e. colimit) of the diagram $$(\dagger)\qquad\ldots\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow \\[-1em] \rightarrow}h_0\sqcup(\sqcup^6h_2)\sqcup h_1\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow} h_0\sqcup h_2\sqcup h_2\sqcup h_1\rightrightarrows h_0\sqcup h_1$$ is $\mathrm{const}_*$, the constant functor $\Lambda_2^2\to\Delta^0$. Here, $h:(\Lambda_2^2)^{\mathrm{op}}\to\mathrm{Spc}$ denotes the Yoneda embedding and 0,1,2 are vertices of $\Lambda_2^2$. This claim implies the desired conclusion since we can write (with notation as in the book by Land and $F:\Lambda_2^2\to\mathrm{Spc}$ given by $X\to Y\leftarrow Z$) $$\operatorname{Map}_{\operatorname{Fun(\Lambda_2^2,\mathrm{Spc})}}(\mathrm{const}_*,F)\simeq\operatorname{Map}_{\mathrm{Spc}}(\Delta^0,F)\simeq\operatorname{Map}_{\mathrm{Spc}}(\Delta^0,\lim F)\simeq\operatorname{Fun}(\Delta^0,\lim F)=\lim F$$ on the one hand and $\operatorname{Map}_{\operatorname{Fun(\Lambda_2^2,\mathrm{Spc})}}(\mathrm{const}_*,F)\simeq\lim_\Delta(*)$ on the other.

To prove that the colimit of $(\dagger)$ is $\mathrm{const}_*$, we can check vertex-by-vertex by HTT. For vertices 0 and 1 it suffices to know that the colimit $K_1$ of the diagram $$\qquad\ldots\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow \\[-1em] \rightarrow}\Delta^0\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow}\Delta^0\rightrightarrows\Delta^0$$ in $\mathrm{Spc}$ (with every map the identity) is weakly contractible, while for vertex 2 we must show that the colimit $K_2$ of the diagram $$\qquad\ldots\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow \\[-1em] \rightarrow}\sqcup^8\Delta^0\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow}\Delta^0\sqcup \Delta^0\sqcup\Delta^0\sqcup\Delta^0\rightrightarrows\Delta^0\sqcup\Delta^0$$ is weakly contractible. We use Corollary 5.5 of this nLab page, which allows us to identify $K_1$ with $\Delta^0$ up to weak equivalence and $K_2$ with $J$ (Joyal interval/nerve of the walking isomorphism) up to weak equivalence. Both simplicial sets are weakly contractible, so we’re done.

The same method allows us to show that we can compute the pullback $X\times_YZ$ via the totalization

$$X\times Z\rightrightarrows X\times Y\times Z\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow} X\times Y\times Y\times Z\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow \\[-1em] \rightarrow}\ldots.$$ It boils down to the same check for the geometric realization $$\qquad\ldots\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow \\[-1em] \rightarrow}h_0\sqcup h_2\sqcup h_2\sqcup h_1\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow} h_0\sqcup h_2\sqcup h_1\rightrightarrows h_0\sqcup h_1.$$ Here the diagrams corresponding to vertices 0 and 1 are the same, while the geometric realization for the vertex 2 corresponds to $\Delta^1$, which is weakly contractible.


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