Fundamental Group of a Quotient Space of $S^{2}$

I have the following problem that I am stuck on.

Let $X$ be the quotient space of $S^{2}=\{\vec{x}\in\mathbb{R}^{3}\mid |\vec{x}|=1\}$ obtained by identifying the points $(0,0,1), (0,0,-1)$, and $(1,0,0)$. Compute the fundamental group $\pi_{1}(X)$.

I know that I should be using the Seifert-van Kampen Theorem, but I'm not sure exactly how to decompose the space. Would the space be homeomorphic to the wedge of three circles? I think that might be right, but I may also be completely wrong. Thanks in advance for any help!

• It's probably easiest to use a cell decomposition of $S^2$ with the three special points as $0$-cells. – Alex Provost Nov 26 '17 at 18:39
• The space is homotopy equivalent to a wedge of two circles and a sphere. To see this, expand the three identified points A,B,C on the sphere into three points with two line segments joining A to C and B to C respectively. Now slide the ends of the two line segments across the sphere until they meet the point C. – Cheerful Parsnip Nov 26 '17 at 18:40

In this case, we let the $0$-cells be the three given points. Note that these all lie on the plane $xz$-plane given by $y = 0$; in this plane, we see there is a short geodesic $a$ connecting $(1,0,0)$ to the north pole and another short geodesic $b$ connecting $(1,0,0)$ to the south pole. These arcs shall be our $1$-cells. Finally, map the unique $2$-cell to the $1$-skeleton in the following manner: consider the circle $S^1 = \{ (x,0,z) : x^2 + z^2 = 1\}$ inside the $xz$-plane. The attaching map of our $2$-cell is given by mapping this circle $S^1$ to the $1$-skeleton via the projection $(x,0,z) \mapsto (\lvert x \rvert,0,z)$. This yields our CW-complex $X$.
Now let $A$ denote the subcomplex consisting of the three $0$-cells, and form the quotient complex $X/A$. This has a single $0$-cell corresponding to the three original points that were identified, and the same pair of $1$-cells and unique $2$-cell as the original space. Thus, letting the basepoint be the $0$-cell, we see that our space is obtained from a wedge of two circles by attaching a $2$-cell along the loop $aa^{-1}bb^{-1} = 1$. This turns out to be trivial, so the normal subgroup $N$ this loops generates is also trivial. We conclude that $\pi_1(X/A) \cong \pi_1(S^1 \vee S^1)/N = \mathbb{Z} * \mathbb{Z}$.