2
$\begingroup$

In homotopy theory we have the Seifert van-Kampen theorem, which is a clean statement about the fundamental groupoid of a pushout in $\mathsf{Top}$. There is also a 2d version of SvK in R Brown's Nonabelian Algebraic Topology book which again is a statement about preserving pushouts in $\mathsf{Top}$. These theorems do not "leak dimension" in the sense that no knowledge of homotopy objects in other dimensions is needed. This is not to say different dimensions are not related - they are in light of the long exact sequence of homotopy groups.

In homology, the big computational tool seems to be the Mayer Vietoris sequence, which relates the homology of pushouts in $\mathsf{Top}$ to the homology of its components. However, this sequence does "leak dimension" - it does requires knowledge of different-dimensional homology objects to compute the homology of a pushout. There is no SvK theorem for homology groups, even in dimensions 1,2 (as far as I know). On the other hand there is no MV sequence for homotopy.

My question is why.

If we step back, MV is a consequence of the long exact sequence in homology and excision, both of which are available in homotopy. So..

Why is there no MV sequence in homotopy? What (absence of) structure in each theory causes this?

Secondly:

Why is there no SvK theorem for homology in any dimension? What structure is missing in homology that enables this theorem in homotopy?

$\endgroup$
  • 2
    $\begingroup$ Seifert-van Kampen 'leaks' dimension since it is about dimensions $0$ and $1$. The fundamental groupoid $\Pi_1(X)$ should be compared to the truncations $t_{\leq n}C_*(X)$ of the (singular) chains, which also preserves homotopy push-outs (both functors preserve homotopy colimits in general). $\endgroup$ – Fernando Muro Mar 12 '15 at 12:09
  • 1
    $\begingroup$ SvK theorem assumes that the spaces are path connected, i.e., $\pi_0=0$; otherwise, $\pi_0$ is also involved. Missing in homotopy groups is the excision theorem. $\endgroup$ – Alex Degtyarev Mar 12 '15 at 13:12
  • $\begingroup$ @AlexDegtyarev what about this? $\endgroup$ – Arrow Mar 12 '15 at 13:23
  • $\begingroup$ What about that? It holds under extra assumptions; under these assumptions, you can write an analog of Mayer--Vietoris as well (Freudenthal's theorem being a typical application). $\endgroup$ – Alex Degtyarev Mar 12 '15 at 13:29
  • 1
    $\begingroup$ @AlexDegty: The groupoid version of the SvKT using a set of base points which I published in 1967 gets over the path-connected assumption - see the presentation referred to in my answer, and mathoverflow.net/questions/40945/… $\endgroup$ – Ronnie Brown Mar 12 '15 at 15:20
1
$\begingroup$

The more useful version of the Seifert-van Kampen theorem deals with the fundamental groupoid $\pi_1(X,A)$ on a set $A$ of base points. The important point here is that groupoids have structure in dimensions $0$ and $1$, and so can handle gluings of (weak) homotopy $1$-types. There is a connectivity condition.

In the next dimension, there are crossed modules over groupoids, which have structure in dimensions $0,1,2$, and they form a category $XMod$. They also model weak homotopy $2$-types, and the 2-dimensional SvKT, proved by Higgins and me, says that the functor $\Pi_2$ from triples to $XMOd$ preserves certain pushouts. Again there is a connectivity condition.

The book, which I will call NAT, referred to in the question, deals in general with a functor $\Pi$ from filtered spaces to crossed complexes, and says that it preserves certain colimits, including pushouts; there is a connectivity condition involved.

In homotopy theory, identifications in low dimensions influence homotopy invariants in high dimensions. A simple example is to form $X=S^n \vee [0,1]$, a sphere with a whisker, which has the same homotopy type as $S^n $. However identifying $0$ and $1$ gives $S^n\vee S^1$, which has a different $\pi_n$. This example is handled by the functor $\Pi$ on filtered spaces mentioned above: see p.259 of NAT.

Some background to these results is given in this presentation.

By contrast, the MV sequence in homology requires no connectivity conditions, but yields only an exact sequence, which determines a term such as $H_n(A \cup B)$ only up to extension, and involves only abelian groups, not modules.

$\endgroup$

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.