# Fundamental group of $X$?

Let $X=X_1\cup X_2\cup X_3$, where $X_1=\{ (x,y,z): x^2 +(y-1)^2+z^2=1\}$ , $X_2=\{ (x,y,z): x^2 +(y+1)^2+z^2=1\}$ and $X_3=\{ (0,y,1): -1\leq y \leq 1 \}$. Find the fundametal group of X.

My guess is, it should be $\mathbb{Z}$. But no idea to prove exactly!

• Are you familiar with Van Kampen? – Jyrki Lahtonen Aug 15 '15 at 22:17
• @Jyrki:I'm familiar with small version of Van Kampen. – Mathsira Aug 15 '15 at 22:19
• What do you mean by the small version of Van Kampen? If you mean the case for only two sets, you could write $X=X_1\cup (X_2\cup X_3)$ and apply the theorem twice. – Michael Burr Aug 15 '15 at 22:22
• Luckily there are fairly nice opens that can be shown to have $X_1, X_2, X_3$ as deformation retracts – JHance Aug 15 '15 at 22:31
• Notice that the sets don't need to be open in $\mathbb{R}^2$, but only open in the induced topology of $X$. So, you can use a slightly larger region in $X$ that retracts onto $X_1$. – Michael Burr Aug 15 '15 at 22:36

Your intuition is correct, $\pi_1(X)\cong\Bbb{Z}$.

With the version of van Kampen that you described it could go as follows.

1. First show that $\pi_1(X_1\cup X_2)$ is trivial. The two spheres kiss each other at the origin. So we can "fatten" both of them to $X_1^+$ and $X_2^+$ by including a small open spherical cap (at the origin) of one of them to the other. We easily see that $X_i^+$ retracts to $X_i$, $i=1,2$. Therefore they have trivial fundamental groups. The intersection $X_1^+\cap X_2^+$ is contractible, so the same applies. The claim follows.
2. Next we define a loop $Y$ that is the union of $X_3$ and the half-meridians from the North poles of the two spheres to the origin. Clearly $\pi_1(Y)=\Bbb{Z}$, because $Y$ is homeomorphic to $S^1$. We fatten $Y$ to $Y^+$by including points on the surfaces of the spheres that are withing $\epsilon$ of the half-meridians. Clearly $Y^+$ retracts to $Y$, so $\pi_1(Y^+)\cong\Bbb{Z}$ as well.
3. In the last step we fatten the union $X_1\cup X_2$ to $X_{12}^+$ by including short segments from $X_3$, say $X_{12}^+$ is $X$ with the middle one-third of $X_3$ thrown away. Clearly $X_{12}^+$ retracts to $X_1\cup X_2$ so its fundamental group is trivial. Furthermore, (draw a picture, if you cannot imagine it in your head) the intersection $X_{12}^+\cap Y^+$ is contractible, so it is simply connected. $X=X_{12}^+\cup Y^+$, so (small) van Kampen gives the claim.
• The purpose of fattening is to make the sets open. Note Michael Burr's comment: Open as subsets of $X$ - not necessarily open as subsets of $\Bbb{R}^3$. – Jyrki Lahtonen Aug 16 '15 at 10:26
• Thank you very much! I understood clearly your well written steps. Also got the idea in the Burr's comment. – Mathsira Aug 16 '15 at 13:58

retract the arc connected $(0,-1,1)$ and $(0,0,0)$ $\\$ and the arc connected $(0,1,1)$ and $(0,0,0)$ $\\$ then $$X \sim S^1 \vee S^2 \vee S^2$$

then the fundametal group of X is Z

• Thank you. Could you explain this with more steps ? I want to study your way too. – Mathsira Aug 16 '15 at 14:00
• Sorry i couldn't add the picture yesterday. retract the red line to the origin. – 张良肇 Aug 17 '15 at 8:32