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The van Kampen theorem is:

If $X$ is the union of path-connected open sets $A_{\alpha}$ each containing the basepoint $x_0 \in X$, and if each intersection $A_{\alpha}\cap A_{\beta}$ is path-connected, then the homomorphism $\Phi : *_{\alpha}\pi_1(A_{\alpha}) \to \pi_1(X)$ is surjective.

In most examples I have seen, $A_{\alpha} \cap A_{\beta}$ is a point, (except some complicated examples). Can you find a simple example where $A_{\alpha}\cap A_{\beta}$ is not a point?

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Every connected surface (i.e. topological two-manifold) can be obtained by identifying pairs of oriented edges of some $2n$-sided polygon. One way to calculate the fundamental group of a connected surface is to start with a corresponding polygon $P$ and let $D$ be a disc in the interior of $P$, then set $A_{\alpha} = D$ and $A_{\beta} = P\setminus\{x_0\}$ where $x_0 \in D$. Note that both $A_{\alpha}$ and $A_{\beta}$ are path-connected, and that $A_{\alpha}\cap A_{\beta}$ is a punctured disc, and is therefore path-connected. By applying the Seifert-van Kampen Theorem, you can then calculate the fundamental group of the surface.

An explicit example of this calculation for the torus can be found here.

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  • $\begingroup$ It seems interesting, thank you very much. $\endgroup$ – 6666 Aug 10 '14 at 0:26
  • $\begingroup$ I have a question,in the example, the fundamental group of Aα is Z*Z, and the fundamental group of Aβ is trivial. As the fundamental group of Aβ is trivial, it seem that the fundamental group of Aα∩Aβ is trivial, but the fundamental group of torus is ZxZ, what's wrong with it? $\endgroup$ – 6666 Aug 10 '14 at 1:09
  • $\begingroup$ The fundamental group of $A_{\alpha}\cap A_{\beta}$ is not trivial. $\endgroup$ – Michael Albanese Aug 10 '14 at 1:14
  • $\begingroup$ Then what is it? $\endgroup$ – 6666 Aug 10 '14 at 1:20
  • $\begingroup$ The punctured disc deformation retracts onto a circle, so it has fundamental group $\mathbb{Z}$. $\endgroup$ – Michael Albanese Aug 10 '14 at 1:22
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For $n\ge2$ consider $X=\mathbb{S}^{n}$, $A_{1}=\mathbb{S}^{n}\setminus\{N\}$, and $A_{2}=\mathbb{S}^{n}\setminus\{S\}$ where $N$ and $S$ are the north and south pole respectively. Then $A_{1},A_{2},\text{and}\,A_{1}\cap A_{2}$ are all path connected. The intersection is homotopic to $\mathbb{S}^{n-1}$ and is not a point. The Seifert-van Kampen Theorem applies and I don't think this example is too difficult.

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  • $\begingroup$ Thanks,but in this case we have already known that the fundamental group of X is trivial, so we don't need to use the theorem. Anyway thank you for your answer. $\endgroup$ – 6666 Aug 10 '14 at 0:20
  • $\begingroup$ @facebook-100006493752734 You're welcome. If you need another example you could consider the connected sum of two $n$-dimensional manifolds where $n\ge3$. In this case the Seifert-van Kampen Theorem can be applied to show that the fundamental group of the connected sum is the free product of fundamental groups. The intersection of the open sets will again not be a single point. $\endgroup$ – user71352 Aug 10 '14 at 0:31
  • $\begingroup$ You're welcome. $\endgroup$ – user71352 Aug 10 '14 at 0:48

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