# Fundamental Group of Torus with Axis

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.

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.
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.)
• @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). – Lorenzo Dec 30 '15 at 18:00