I am studying for my comprehensive exams and have come across the following question, which I have been struggling with:

Let $T$ be the torus given by rotating the circle $\{(x,0,z)\in\mathbb{R}^3\mid(x-2)^2+z^2=1\}$ around the $z$-axis, and let $X$ be the union of $T$ and the $x$-axis. Using the Seifert-van Kampen Theorem (or otherwise), find the fundamental group $\pi_1(X)$.

Well, we know that the space $X$ deformation retracts to the Torus union the interval $[-3,3]$ (on the $x$-axis), so they have isomorphic fundamental groups. To apply SVK, I imagine we will want $U$ to be the torus (with some fringe of the $x$-axis to make it open), but I can't find a good choice of $V$ that makes things work out. If anyone has suggestions, it would be greatly appreciated.


1 Answer 1


Lemma: Suppose $X$ is obtained from $Y$ by attaching a cell $D$ by a map $f : \partial D \to Y$, and $X'$ is obtained by the map $f'$. If $f$ and $f'$ are homotopic, then $X'$ and $X$ are homotopy equivalent. In particular, they have the isomorphic fundamental groups.

Reference: Prove homotopic attaching maps give homotopy equivalent spaces by attaching a cell

In this case, first think of your space as a torus union an interval $[-3,3]$ on the x-axis. We mentally break this interval up into three parts, namely three intervals glued onto the torus at four points. For each attached interval individually you can homotope it attaching map around freely using the lemma above (the cell is the interval). So move them so that what you see is a torus with a bouquet of three circles wedged on. Now apply Van Kampens (more specifically its corollary, the fundamental group of a wedge.)

If you insist on using Van Kampen's:

You can divide the torus in half around the x-z plane. The intersection will be two theta shapes connected by a line. This is clearly path connected, and you can enlarge your open sets a little in order to apply SVK. You can compute the fundamental group of each half by collapsing the torus part. Presumably this computation works out, but it seems pretty tedious - at some point you will have to verify a commutator relationship, since you are going to have the fundamental group of the torus in there as part of a free product (by the first computation).

  • $\begingroup$ OK, I can see why the question included the (or otherwise) clause. Homotoping those segments of the axis is a great idea! So if we take U to be the torus and V to be the wedge (both with some open fringe) then would UnV deformation retract to a point so that pi1(X)=pi1(U)*pi1(V)=(ZxZ)*(ZZZ)? Also, is there a way to write that group better? $\endgroup$ Dec 30, 2015 at 4:35
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    $\begingroup$ @LoganTatham That's what I got too. I would write $(Z \times Z) * (F_3)$ (not that it makes a difference conceptually). I doubt that there is a way to simplify that description, describing at as as a free product tells you exactly what it is (what the elements are and how to multiply in it). $\endgroup$
    – Elle Najt
    Dec 30, 2015 at 18:00
  • $\begingroup$ There are lots of other questions on old qualifying exams that I'm studying that use a very similar strategy. This has been very helpful! $\endgroup$ Dec 31, 2015 at 1:02

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