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I have been reading Jacob Lurie's book "Higher Algebra", version May 8, 2011. One is grateful to him for covering such a lot of ground and for making it all so readily available.

My attention was drawn to the Section A.3 on "The Seifert-van Kampen Theorem" p. 845.

It starts by stating the classical theorem determining the fundamental group of pointed space which is a union of two open sets with path-connected intersection. (The most general theorem of this type is for the fundamental groupoid on a set $A$ of base points for a space $X$ which is the union of a family $\mathcal U$ of open sets and such that $A$ meets each path-component of all 1-,2-,3-fold intersections of the sets of $\mathcal U$).

It states: " In this section, we will prove a generalization of the Seifert-van Kampen theorem, which describes the entire weak homotopy type of $X$ in terms of any sufficiently nice covering of X by open sets: Theorem A.3.1." However this Theorem makes no mention of groups or connectivity conditions.

So my question is: How does one deduce the SvKT as stated there, or its more general version, from Theorem A.3.1?

Theorem A.3.1 itself seems closely related to classical theorems on excision, showing the singular complex is chain homotopy equivalent to the singular complex of $\mathcal U$-small simplices. (I don't have the earliest reference for this, but I like the proof by R. Sch\"on from Proc. AMS 59 (1976).)

A particular point for the deduction of the most general version of the SvKT is: why the number 3? One explanation is that it has to do with the Lebesgue dimension of $\mathbb R^2$.

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My guess would be that 3 is 1+2, where 1 is the dimension of the groupoids in question (the classical SvKT is about 1-groupoids) and 2 is a universal constant for $n$-colimits --- a colimit of $n$-groupoids can be calculated using coproducts and realizations of truncated simplicial objects (generalized coequalizers) with $n+2$ terms. If we were calculating the fundamental 0-groupoid $\pi_0$, we should only need 1- and 2-fold intersections, and similarly for fundamental $n$-groupoids we should expect to need up to $(n+2)$-fold intersections. –  Mike Shulman Jul 15 '12 at 21:17
    
That is a reasonable suggestion, and concurs with +1 to deal with homotopies (for the strict structures of our fundamental higher groupoids), and another +1 to deal with the covering dimension. Of course Higgins and I have good reasons for using cubical rather than simplicial methods. Major features of the two strands of work are (i) that they deal with structured spaces (filtered or $n$-cubes), and (ii) that the construction of the corresponding "fundamental" object, essential to the work, is not straightforward. –  Ronnie Brown Jul 16 '12 at 11:05
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2 Answers 2

I don't actually see how to deduce the version of the classical SvKT on a set of base points $A$ directly from Lurie's version. It seems that to apply Lurie's theorem, we would need the stronger hypothesis that $A$ meets every path component of every finite intersection of open sets in $\mathcal{U}$ (which is reasonable, since the conclusion of Lurie's theorem is a stronger statement about fundamental $\infty$-groupoids rather than just fundamental 1-groupoids). But I think we can adapt his proof to derive the classical version.

Here's the first thing I'm going to try to prove: let $\chi:C\to \mathcal{O}(X)$ be a functor, with $C$ a small category and $\mathcal{O}(X)$ the poset of opens in $X$. For each $x\in X$, define $C_x$, as Lurie does, to be the full subcategory of $C$ spanned by those objects $c$ with $x\in\chi(c)$. Assume that for every $x$, the nerve of $C_x$ is simply connected. Then we have $\Pi_1(X) \cong \mathrm{colim}_{c\in C}\; \Pi_1(\chi(c))$, the colimit being a weak 2-colimit of groupoids.

Mimicking Lurie's argument in the 1-truncated case, we have a (pseudo 2-)functor $F:\mathrm{Gpd}^{\mathcal{O}(X)^{\mathrm{op}}} \to \mathrm{Gpd}$ defined by Kan extension from the functor $\Pi_1 : \mathcal{O}(X) \to \mathrm{Gpd}$. The (2,1)-topos $\mathrm{Sh}_{(2,1)}(X)$ is the (bicategorical) localization of $\mathrm{Gpd}^{\mathcal{O}(X)^{\mathrm{op}}}$ at the covering sieves, and Lurie's A.3.2 shows that $F$ inverts these covering sieves and hence factors through $\mathrm{Sh}_{(2,1)}(X)$. In particular, this induced functor $F:\mathrm{Sh}_{(2,1)}(X) \to \mathrm{Gpd}$ preserves (bicategorical) colimits.

Thus, it suffices to show that our functor $\chi:C\to \mathcal{O}(X)$ has colimit $X$ (the terminal object) when composed with the Yoneda embedding into $\mathrm{Sh}_{(2,1)}(X)$. And since $\mathrm{Sh}_{(2,1)}(X)$ has enough points (being sheaves on a topological space), it suffices to check this on all stalks. (At a finite categorical dimension, there is no hyper-incompleteness to worry about.) But at the stalk over $x\in X$, the $C$-diagram is trivial at those $c\in C_x$ and empty at the others, so its colimit is simply the groupoid reflection of $C_x$, which was assumed to be terminal (this is equivalent to the nerve of $C_x$ being simply connected).

This completes the proof of the 1-groupoidal version of Lurie's theorem. Now let's deduce a more classical statement. Let $X$ be our space and $\mathcal{U}$ an open cover of it. Define $C$ to be the category of 1-, 2-, or 3-fold intersections of open sets in $\mathcal{U}$, whose morphisms are the canonical inclusions from an $(n+k)$-fold intersection to an $n$-fold intersection, and let $\chi$ be the obvious functor.

For any $x\in X$, the category $C_x$ is obviously nonempty (because $\mathcal{U}$ covers $X$) and connected (because if $x\in U$ and $x\in V$, then $x\in U\cap V$). It is not much harder to see that it is simply connected: any two parallel zigzags of inclusions can be made equal by passing through at most triple intersections. Thus, we have

$$\Pi_1(X) \cong \mathrm{colim}_{c\in C} \; \Pi_1(\chi(c)).$$

Now let $A\subseteq X$ be a subset which meets all path components of all 1-, 2-, and 3-fold intersections of open sets in $\mathcal{U}$. Then $\Pi_1(\chi(c))$ is equivalent to its full sub-groupoid $\Pi_1(\chi(c),A)$ spanned by objects that are points of $A$, as is $\Pi_1(X)$. Since 2-dimensional colimits are invariant under equivalence of groupoids, the above statement passes to these groupoids as well.

Now the generalized SvKT of Ronnie and his coauthors amounts to asking that $\Pi_1(X)$ be the strict colimit of the functor $c\mapsto \Pi_1(\chi(c),A)$, in the 1-category of groupoids, so it basically suffices to show that this strict 1-colimit is also a (weak) 2-colimit. Now $C$ is a direct category, and $\mathrm{Gpd}$ is a model category with the canonical model structure (weak equivalences are equivalences, cofibrations are injective on objects), so $\mathrm{Gpd}^C$ inherits a Reedy model structure for which the adjunction

$$ \mathrm{colim} : \mathrm{Gpd}^C \;\rightleftarrows\; \mathrm{Gpd} : \Delta $$

is Quillen. It follows that the 1-colimit of a Reedy cofibrant diagram is also a 2-colimit. Unfortunately, $c\mapsto \Pi_1(\chi(c),A)$ is not Reedy cofibrant, but it is "partly" so. For instance, for any $U\in \mathcal{U}$, consider the latching object

$$ L_U = \mathrm{coeq}\left( \coprod_{V,W} \Pi_1(U\cap V\cap W,A) \;\rightrightarrows\; \coprod_{V}\Pi_1(U\cap V,A) \right) $$

Then the map $L_U \to \Pi_1(U,A)$ is injective on objects; thus our functor is at least Reedy cofibrant "at the top level". It will suffice to show that if $G\in \mathrm{Gpd}^C$ is "sufficiently Reedy cofibrant" in senses like this, then $\mathrm{colim}(G)$ can be calculated in a homotopy-invariant way.

Let $G\in \mathrm{Gpd}^C$, and enumerate the elements of $\mathcal{U}$ (perhaps transfinitely) as $(U_\alpha)_{\alpha<\lambda}$. We will define a transfinite sequence of groupoids $$ H_0 \to H_1 \to H_2 \to \cdots $$ such that $H_\alpha$ is the colimit of $G$ restricted to the subcategory of $C$ determined by the $U_\beta$ with $\beta<\alpha$, and their pairwise and triple intersections. Of course we can take $H_0 = 0$. Now given $H_\alpha$, define $H_{\alpha+1}$ to be the pushout of $H_\alpha$ and $G(U_{\alpha})$ along $$ K_\alpha = \mathrm{coeq}\left( \coprod_{\beta,\gamma < \alpha} G(U_\alpha \cap U_\beta \cap U_\gamma) \;\rightrightarrows\; \coprod_{\beta<\alpha} G(U_\alpha \cap U_\beta) \right). $$ For limit $\alpha$, we of course define $H_\alpha$ to be the colimit. It is easy to verify that each $H_\alpha$ is the colimit as asserted, and thus $\mathrm{colim}_{\alpha<\lambda} \; H_\alpha = \mathrm{colim} \; G$.

Now suppose $G$ has the property that the map $K_\alpha \to G(U_\alpha)$ is injective on objects (a cofibration) for every $\alpha$. Then the pushout defining $H_{\alpha+1}$ is a homotopy pushout and thus homotopy-invariant. Moreover, the map $H_\alpha \to H_{\alpha+1}$ is again a cofibration, so the colimits at limit stages are also homotopy colimits and thus homotopy-invariant. Therefore, for $G$ with this property, strict colimits are 2-colimits. But our functor $c\mapsto \Pi_1(\chi(c),A)$ does have this property (it is a slightly stronger version of being "Reedy cofibrant at the top level").

Thus, its strict colimit is also a homotopy colimit, so $\Pi_1(X,A)$ is equivalent to this strict colimit. The theorem of Ronnie and coauthors asserts that it is in fact isomorphic to this strict colimit, but it is easy to check that it has the same set of objects as the strict colimit (namely $A$), and an equivalence of groupoids which is bijective on objects is an isomorphism.

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Bravo! Nice to see this sorted out. –  David Roberts Jul 16 '12 at 23:06
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Mike's seems a good complete answer to my question, though I think beginners in algebraic topology might prefer the proof of the general result in our paper Archiv. Math. 42 (1984) 85-88! Part of my confusion behind my question is the distinction between "A generalises B", and "A implies B", especially if A also implies excision, MV, .... Mike's proof leaves open the question of proofs of the 2-d and higher dimensional results: the current proofs involve new methods, but the theorems directly imply classical results in algebraic topology, e.g. on $\pi_n(S^n)$, and on free crossed modules. –  Ronnie Brown Jul 17 '12 at 10:56
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I should recall that in the first 1968 edition of my book now "Topology and Groupoids", the argument that a retract of a pushout is a pushout was used there to compute $\pi_1(S^1)$ perhaps prompting the then MAA reviewer to write: "This reads like a book on topology written by a category theorist." Another local-to-global result in the current edition, and as tricky as the SvKT, computes $\pi_1(X/G)$ as $(\pi_1X)//G$ (orbit groupoid) for certain useful situations. Is there a generalisation of this result to higher dimensions? or other generalisations? –  Ronnie Brown Jul 17 '12 at 16:16
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Also, personally, when I say or hear "A generalizes B", I generally assume that this means in particular that "A implies B". In this sense, Lurie's theorem is not strictly speaking a generalization of the most general possible SvKT with a set of basepoints (at least, I don't see how). However, in the text surrounding the theorem, he only claims that it generalizes the most classical SvKT about unions of two open subsets, and this does seem to be true. –  Mike Shulman Jul 17 '12 at 17:34
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Finally, I agree, I would very much like to know whether higher-dimensional strict versions can be extracted from Lurie's theorem. The first steps of the proof I gave should work just as well for arbitrary $n>1$, but it seems that the identification of weak and strict colimits of $n$-groupoids may fail because we don't have a good model structure on algebraic $n$-groupoids, and if we did then its cofibrations would probably be hard to understand. Maybe your cubical approaches could be rephrased as a solution to this problem? –  Mike Shulman Jul 17 '12 at 17:38
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My impression (which is that of a total non-expert) is the following:

Spaces, according to the "homotopy hypothesis," are the same thing as $(\infty, 0)$-categories. This comes out of Lurie's (and Joyal's) work if one accepts that quasi-categories are an appropriate model for $(\infty, 1)$-categories: it's a result that a quasi-category with every 1-morphism invertible is the same thing as a Kan complex.

Let $X$ be a space, and let $U, V$ be open subsets which cover $X$. Then we have a homotopy pushout diagram expressing $X$ as the homotopy pushout $U \sqcup_{U \cap V} V$. In other words, if we think in terms of higher groupoids, $X$ is the homotopy pushout of the $\infty$-groupoids corresponding to $U, V, U \cap V$.

Describing the homotopy type of a homotopy pushout completely (i.e., in terms of homotopy groups and such) is really hard: otherwise we would know all the homotopy groups of spheres! What I take from this is that explicit models for higher category theory that one can easily compute with are likely to be very complicated, or otherwise homotopy theory would become easy.

However, there might be more luck if we restrict to special cases. For instance, there is an equivalence between 1-truncated spaces (spaces with no higher homotopy groups than $\pi_1$) and groupoids, given by taking the fundamental groupoid. As it happens, we can compute with groupoids: there is, for instance, a nice model category presentation of groupoids, and we can work out what the homotopy pushout of groupoids is.

Since truncation below $n$ is a left adjoint, this then amounts to saying that taking the fundamental groupoid sends homotopy pushout squares in spaces to homotopy pushout squares in groupoids. This is precisely the classical van Kampen theorem (stated for groupoids rather than groups, though.) More generally, we can say that the "fundamental $n$-groupoid of a space" (by which I mean the truncation below $n$) commutes with homotopy push-outs.

(Example: if we want to show that $\pi_1(S^1) = \mathbb{Z}$, we observe that $S^1 $ is a homotopy pushout $\ast_{\sqcup S^0} \ast$, so we have to compute the "suspension" of the discrete groupoid on two elements. Taking the homotopy pushout amounts to adding two isomorphisms identifying the two points, which gives a groupoid equivalent to $\mathbb{Z}$.)

I understand that you have done work generalizing the classical van Kampen theorem to higher homotopy groups. My guess that, in Lurie's language, some of that would translate into the construction of explicit algebraic model for 2-and-higher-groupoids (as opposed to 2-truncated spaces) and a means of computing homotopy pushouts (but I'm only speculating here), i.e., it seems to me that it would be quite different from what "Higher Algebra" does.

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The subtlety is that the work of Ronnie and collaborators deals entirely with colimits, but Lurie's A.3.1 is a homotopy colimit. Groupoids still form a 2-category, so the truncation functor would, I expect, recover the pushout as a 2-colimit. The question is more about recovering the exact statement of the classical SvKT from the version with singular simplices, in particular, how one gets the necessary assumptions for the SvKT to hold, which are tight, from the pretty much assumptionless A.3.1 –  David Roberts Jul 16 '12 at 0:47
    
I would like to emphasise that my work on higher homotopy SvKTs deals with strict colimits of strict structures: the work with Philip Higgins enables (some!) complete determinations of relative homotopy groups as modules (crossed if $n=2$) over fundamental groupoids, including nonabelian second relative homotopy groups; the work with Jean-Louis Loday enables (some!) complete determination of $n$-adic homotopy groups, which are also nonabelian at the bottom level. The proofs do use (strict) higher homotopy groupoids; all may be quite distinct from the aims of "Higher Algebra". –  Ronnie Brown Jul 16 '12 at 9:20
    
My apologies if my speculations were way off. Your work is definitely on my to-read list, but I haven't yet gotten there... –  Akhil Mathew Jul 16 '12 at 11:07
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@Ronnie: The connectivity conditions are just to guarantee that the fundamental groupoids of $U$, $V$, and $U\cap V$ are equivalent to the fundamental groups on the base point (viewed as groupoids with one object). Since homotopy pushouts preserve equivalences, the pushout of these groups is the fundamental group of $U\cup V$. Leaving out the connectivity conditions just gives a more powerful result at no additional cost. –  Marc Hoyois Jul 16 '12 at 18:58
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Akhil computes the fundamental group of $S^1$ without using any set of base points. Once you know that such colimits (of full fundamental groupoids) are homotopy colimits, choosing a set of base points that intersects every open in your diagram only modifies the groupoids in the diagram (and hence their colimit) up to equivalence, so it's not a useful operation! As Mike remarks in his answer, the point of 3-fold intersections is that we can disregard 4-fold intersections and higher without changing the homotopy colimit (this much should remain true for higher $n$, replacing $3$ by $n+2$). –  Marc Hoyois Jul 16 '12 at 23:05
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