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!
 A: Your intuition is correct, $\pi_1(X)\cong\Bbb{Z}$.
With the version of van Kampen that you described it could go as follows.


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*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.

*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.

*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.

A: 
retract the arc connected  $(0,-1,1)$ and $(0,0,0)$
$ \\ $
and the arc connected  $(0,1,1)$ and $(0,0,0)$
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then
$$ X \sim  S^1 \vee S^2 \vee S^2 $$
then the fundametal group of  X is Z
