Fundamental group of two spheres and circle

Question

Let two $S^2$, $X_1 = S^2, X_2=S^2$ and $X_3 = S^1$. Let $N_i,S_i$ poles of $X_i$, $i = 1,2$ and $n,s$ poles of $X_3=S^1$. Consider the space that $S^1$ is between the two spheres and glued like $N_1 \sim n \sim N_2$ and $S_1 \sim s \sim S_2$. Im triying to compute the fundamental group but the intersection of the open sets is a $S^1$ and I cant compute the amalgamed product. Any idea?

Edit: I extend my idea but I think its wrong. Consider $U$ and $V$ open that are homotopy eq. to $S^2 \vee S^1 \vee S^1$. $U \cap V$ is homotopy equ. to $S^1$. Then $i_*,j_* : \mathbb{Z}\to \mathbb{Z}\star \mathbb{Z}$ are the identity map. Then $i_*(b) =b$ and $j_*(c) = c$, where $b$ and $c$ is a generator of $\pi_1U$ and $\pi_1V$. So

$$ \pi_1() = \langle a,b,c,d : i_*(b) = j_*(c)\rangle = F(a,b,d). $$

Answer 1

I believe your result is correct. It's just that your argument seems kind of vague and I don't know your definition of $U$ and $V$.

Visualizing $X$ as(please forgive my poor drawing):

and then cut it vertically in the left and right of the red circle($X_3$), and each remove some part from $X_3$, I got $U$ and $V$:

"Squashing" these "broked" sphere, $U$ and $V$ are homotopic to:

the fundamental group of both spaces are the free group with two generators. And $U\cap V$ is:

$U\cap V$ is homotopic to two circles attaching poles together and the generator gives a relation that a generator in $\pi_1(U)$ is equal to a generator in $\pi_1(V)$ except all other trivial relations. Hence $\pi_1(X)$ is a free group with 3 generators.

I posted the answer one day later and I realize there is a much easier way to calculate it.

(Method 2) By using van kampen theorem and homotopy equivalence twice:

(Method 3) Or using Hatcher's Algebraic Topology Proposition 1.26(Page 50), the space $X$ is just attaching two 2-cells to a 1-skeleton containing 3 1-cells.

Using the proposition(taking from Hatcher's Algebraic Topology), it's a free group with 3 generators.

Answer 2

This is a little sketchy, but...

Call your overall space $X$, and take a "longitude" line $L$ that goes from $N_1$ to $S_1$ in $X_1$. Now gradually shrink $L$ to a point. This appears to give a homotopy equivalence from $X$ to the wedge of three spaces: $X_1 / L$, $X_2 / N_2 \sim S_2$, and $S^1 / n \sim s$. So we can use van Kampen and take the free product of the fundamental groups of these three spaces.

$X_1 / L$ is a sphere, so its fundamental group is trivial.

$X_2 / N_2 \sim S_2$ appears to have fundamental group $\Bbb Z$. If we imagine it as a skinny cylinder with both ends collapsed together to the basepoint, the only thing that matters about a loop is how many times it passes through the basepoint and in which direction.

$S^1 / n \sim s$ is a bouquet of two circles, so its fundamental group is $\Bbb Z * \Bbb Z$.

So I get $\Bbb Z * \Bbb Z * \Bbb Z$.