5
$\begingroup$

The Seifert-van Kampen theorem is the classical theorem of algebraic topology that the fundamental group functor $\pi_1$ preserves pushouts; more often than not this is referred to simply as the van Kampen theorem, with no Seifert attached. Curious as to why, I tried looking up the history of the theorem, and (in the few sources at my immediate disposal) could only find mention of van Kampen; the wikipedia page of Herbert Seifert doesn't even mention the theorem.

So my question is: Why does the theorem bear Seifert's name?

$\endgroup$
2
  • 1
    $\begingroup$ Seifert's name is attached to many things already, so it's less distinctive. If we called it Seifert's theorem we'd always have to say something like "Seifert's fundamental group theorem". But saying Van Kampen is easier, since we don't need to worry people will misunderstand what theorem we're referring to. $\endgroup$ May 9 '14 at 12:32
  • 1
    $\begingroup$ Based on the answer below, a natural question seems to be "what was van Kampen's contribution" since the reference given is solo-authored by Seifert. $\endgroup$ Aug 25 '16 at 19:59
6
$\begingroup$

According to Jahrbuch der Mathematik this theorem is in:

  • Seifert, H. Konstruktion dreidimensionaler geschlossener Räume. (German) JFM 57.0723.01 Berichte Leipzig 83, 26-66. Technische Hochschule Dresden, Diss (1931).

See the (long) review in zbMATH by Erika Pannwitz. The relevant sentence is:

  • (Die erforderlichen Hilfsmittel werden in § 2 und 3 der Arbeit zusammengestellt; § 3 enthält insbesondere Aussagen über die Fundamentalgruppe eines Komplexes, der aus zwei Komplexen mit zusammenhängendem Durchschnitt zusammengesetzt ist.)

See also footnote 65 on page 19 of:

$\endgroup$
3
  • 1
    $\begingroup$ The paper Gramain, A. Le th\'eor`eme de van Kampen. Cah. Top. G\'eom. Diff. Cat. 33 (1992) 237--251, gives some more history. Actually van Kampen's paper is quite difficult to follow, and the matter was clarified by Crowell to the modern version. It seemed to me in 1965 that there was an anomaly that this theorem did not calculate the fundamental group of the circle, THE basic example in algebraic topology. This led to the use of groupoids in this area. See the discussion at mathoverflow.net/questions/40945/… $\endgroup$ May 9 '14 at 9:58
  • $\begingroup$ @Ronnie: The paper of Gramain looks very informative. I seems that Seifert treated only the case of 2 sub complexes of a simplicial complex with connected (from the review I cited above) intersection. $\endgroup$ May 9 '14 at 11:16
  • $\begingroup$ Wrt the paper by Seifert, that is also my understanding. The proof by Crowell is modern in that it goes by verifying the universal property; this proof generalises easily to the many base point case, i.e. non connected spaces, so dealing with many not unusual examples, including of course the circle; the proof also extends, with new ideas, to higher dimensions (see our EMS Tract vol 15, 2011, "Nonabelian algebraic topology"). $\endgroup$ May 9 '14 at 14:17
4
$\begingroup$

Since Seifert's article is hard to come by, here is the relevant excerpt (pp. 33f):


$K'$ und $K''$ seien zwei zusammenhängende Teilkomplexe eines zusammenhängenden $n$-dimensionalen Komplexes $K$, so daß jedes $i$-dimensionale Element von $K$ ($i=0,1,\dots ,n$) mindestens einem der Komplexe $K'$ und $K''$ angehört. Der Durchschnitt $D$ von $K'$ und $K''$, der sicher nicht leer ist, sei ebenfalls zusammenhängend. $\mathcal{K},\mathcal{K}',\mathcal{K}'',\mathcal{D}$ seien die Fundamentalgruppen von bzw. $K,K',K'',D$. Wählt man den Anfangspunkt für die geschlossenen Wege in $D$, so ist jeder fundamentale Weg von $D$ zugleich fundamentaler Weg von $K'$ und $K''$. Somit entspricht jedem Element von $\mathcal{D}$ ein Element von $\mathcal{K'}$ und ebenso ein Element von $\mathcal{K''}$. Dann gilt

Satz 1: $\mathcal{K}$ ist eine Quotientengruppe des freien Produktes $\mathcal{K'}\cdot \mathcal{K''}$ von $\mathcal{K'}$ und $\mathcal{K''}$. Man erhält $\mathcal{K}$ aus $\mathcal{K'}\cdot \mathcal{K''}$, wenn man je zwei Elemente von $\mathcal{K'}$ und $\mathcal{K''}$, die demselben Element von $\mathcal{D}$ entsprechen, zusammenfallen lässt.


For convenience, here it is translated (by resisting the temptation to use current terminology and notation, I have attempted to preserve the perspective of the era in which the article was written):

Let $K'$ and $K''$ be two connected subcomplexes of a connected $n$-dimensional complex $K$, so that every $i$-dimensional element of $K$ ($i=0,1,\dots ,n$) belongs to at least one of the complexes $K'$ and $K''$. Assume that the intersection $D$ of $K'$ and $K''$, which is certainly not empty, is also connected. Let $\mathcal{K},\mathcal{K}',\mathcal{K}'',\mathcal{D}$ denote the fundamental groups $K,K',K'',D$, respectively. If we choose a base point for the closed paths in $D$, then every fundamental path in $D$ is at the same time a fundamental path in $K'$ and in $K''$. Therefore, every element of $\mathcal{D}$ corresponds to an element von $\mathcal{K'}$ and also an element of $\mathcal{K''}$. Then the following hold

Theorem 1: $\mathcal{K}$ is a quotient group of the free product $\mathcal{K'}\cdot \mathcal{K''}$ of $\mathcal{K'}$ and $\mathcal{K''}$. One obtains $\mathcal{K}$ from $\mathcal{K'}\cdot \mathcal{K''}$, by identifying any two elements of $\mathcal{K'}$ and $\mathcal{K''}$, which correspond to the same element of $\mathcal{D}$.

$\endgroup$

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.