# Does the Seifert-van Kampen Theorem applied to loop spaces say anything about higher homotopy groups?

The Seifert-van Kampen tells us that $\pi_1$ preserves pushouts of the form $U \cap V \subseteq U, V$ (for $U \cap V$, $U$, $V$ open, path-connected).

Can this information be used to say anything about higher homotopy groups, eg. $\pi_2$ using the fact that $\pi_2(X,x)=\pi_1(\Omega X, \mathrm{const}\ x)$?

• I mean, there's a tautological answer, which is that if you find $U, V$ satisfying the right conditions in $\Omega X$ then you can apply Seifert-van Kampen to it. – Qiaochu Yuan Sep 30 '14 at 4:09
• A limitation on how strong the statements you can make is the following: Seifert-van Kampen implies that the fundamental group of a finite CW complex is finitely presented. This is false for $\pi_2$. – Qiaochu Yuan Sep 30 '14 at 4:10
• @QiaochuYuan, yeah, but has anyone used this (your first comment) to compute anything? – user40167 Sep 30 '14 at 4:18
• Probably not; it's pretty hard to work with loop spaces that directly. A standard method of computing $\pi_2$ is to use the fact that it's naturally isomorphic to $H_2$ of the universal cover, which is usually much more tractable to work with. – Qiaochu Yuan Sep 30 '14 at 4:29

There are extensions of the Seifert-van Kampen Theorem to higher dimensions, but these use a new approach. The background to this is given in a talk I gave at the IHP in Paris in June, and which is available on my preprint page. Full details of most of that talk are in the EMS Tract Vol 15, (2011) Nonabelian Algebraic Topology (pdf available), and another survey is here.

The general idea is that one needs nonabelian algebraic models of homotopy theory to obtain colimit theorems, and for these one needs "structured spaces", for example (i) filtered spaces, or (ii) $n$-cubes of spaces. Initially, the first gives some information on relative homotopy groups, and the second gives information on $n$-adic homotopy groups. From this one has to extract information on homotopy groups, and this may not be easy.

As an example for the relative case, the book cited gives computations of the homotopy $2$-type of some mapping cones of $Bf: BG \to BP$, for groups $G,P$; the answer is a crossed module of the form $\mu: M \to P$ but the second homotopy group is Ker $\mu$.

As an example for the triadic case, the $3$-type of the suspension $X=SK(G,1)$ is calculated in terms of a "commutator morphism" $\kappa: G \otimes G \to G$ and $\pi_3(X) \cong$ Ker $\kappa$. A bibliography on this "nonabelian tensor product" now has 131 items, and lots of computations have been done, see item [117], with interest especially from group theorists, because of the commutator connection.

So we were really forced "out of the box" to get these results. The initial results on second relative homotopy groups were published in 1978.

October 1, 2014: The short answer to your question is that the approach will not give new information, and I was trying to show that there are other versions which do give useful information.

The reason why the suggestion does not work is as follows. A path $g: I \to \Omega (X.x_0)$ in a loop space on $X$ can also be regarded as a map $f:I^2 \to X$ of a certain kind. If $X$ is the union of two open sets, then the inverse image of these gives an open cover of $I^2$ by sets which may not be connected. One proof of the SvKT goes by subdividing a path into pieces mapped into the sets of the cover. In the suggested case, we need to take not just a subdivision of the path $g$ but a subdivision of the map $f$ into little squares; the fundamental group involves composition of paths or loops, but the proposed idea needs compositions of squares, and in 2 directions. So one needs new concepts to make it work, and then one can get new information.

The nice thing about the SvKT is that it gives nonabelian colimit information, but in dimension 1. To get analogues in higher dimension needs new ideas.

Hope that helps.

• I don't see how this is at all relevant to my question – user40167 Sep 30 '14 at 16:24
• People talk about "the Seifert-van Kampen Theorem" but are often unaware that there are generalisations to groupoids in dimension 1 and to higher dimensions. So I am pointing out that these exist. – Ronnie Brown Nov 6 '14 at 18:24
• The traditional fundamental group methods are inadequate for higher homotopy groups directly; the usual SvK Theorem, with a single base point, does not even give the fundamental group of the circle! There are presentations on my preprint page which explain how to go further. – Ronnie Brown Aug 1 '17 at 21:17

Here is an elaboration on the comment I made. Suppose $A, B, X = A \cup B, A \cap B$ satisfy both the conditions of Seifert-van Kampen and the conditions for the Mayer-Vietoris sequence

$$\dots \to \widetilde{H_{n+1}}(X) \to \widetilde{H_n}(A \cap B) \to \widetilde{H_n}(A) \oplus \widetilde{H_n}(B) \to \widetilde{H_n}(X) \to \dots$$

to exist. If in addition $A, B$, and $A \cap B$ are all $(n-1)$-connected, $n \ge 2$, then by Seifert-van Kampen $\pi_1(X)$ vanishes, and by Mayer-Vietoris and Hurewicz, $X$ is also $(n-1)$-connected. Hence $H_n \cong \pi_n$ for all four spaces and Mayer-Vietoris gives that $\pi_n(X)$ fits into an exact sequence

$$\dots \to H_{n+1}(X) \to \pi_n(A \cap B) \to \pi_n(A) \oplus \pi_n(B) \to \pi_n(X) \to 0.$$

A generalization of this result (I think) is given by the Blakers-Massey theorem. Another important way to compute higher homotopy groups is given by the long exact sequence of a fibration.

Ok, so here's a thought.

EDIT: this doesn't work, as $\Omega(U \cup V) \neq \Omega(U) \cup \Omega(V)$

Let $U, V$ open subsets of $U \cup V$. Then $\Omega U$, $\Omega V$ are open in $\Omega (U \cup V)$ in the compact-open topology. To apply Seifert-van Kampen, we need $\pi_0(U) = \pi_0(V) = \pi_0(U \cap V)$ trivial. That is, we require $U$, $V$, $U \cap V$ to be simply connected.

The conclusion being that if $U$, $V$, $U \cup V$ are $n$-connected, and open in $U \cup V$, then $\pi_{n+1}$ preserves the pushout?

• @ krey: The groupoid version of the Seifert-van Kampen Theorem does not require these connectivity assumptions. However see my answer below. – Ronnie Brown Oct 1 '14 at 14:30
• @krey: a version of this statement is true and can be deduced from Mayer-Vietoris together with the Hurewicz isomorphism. The problem with the argument you're trying to run is that it's not true that $\Omega(U \cup V) = \Omega(U) \cup \Omega(V)$, which is of course another condition you'd need to apply Seifert-van Kampen. – Qiaochu Yuan Nov 3 '14 at 1:47
• @QiaochuYuan, wow! thanks for pointing this out – user40167 Nov 3 '14 at 1:55