I have this exercise:
Compute the fundamental group $\pi_1(X)$ of the space $X=S^2 \cup \{ (x,0,0) : x\in [-1,1]
\} \cup \{(0,y,0):y\in [-1,1] \} \cup \{(0,0,z):z\in [1,1] \}$.
I tried that via Seifert-Van Kampen:
Let $I_x = \{ (x,0,0) : x\in [-1,1]
\}$, $I_y = \{(0,y,0):y\in [-1,1] \}$, $I_z = \{(0,0,z):z\in [1,1] \}$.
Then $\pi_1(X) = \pi_1(S^2 \cup I_x \cup I_y \cup I_z) = \pi_1((S^2 \cup I_x) \cup (I_y \cup I_z)) \cong \pi_1(S^2 \cup I_x) \ast_{\pi_1((S^2 \cup I_x) \cap (I_y \cup I_z))} \pi_1(I_y \cup I_z) \cong (\pi_1(S^2) \ast_{\pi_1(S^2\cap I_x)} \pi_1(I_x)) \ast_{\pi_1((S^2 \cup I_x) \cap (I_y \cup I_z))} (\pi_1(I_y) \ast_{I_y\cap I_z} \pi_1(I_z)) \cong (\{0 \} \ast_{\pi_1(S^0)} \{ 0 \}) \ast_{\pi_1((0,0,0))} (\{0 \} \ast_{\pi_1((0,0,0))} \{ 0 \})=(\{0\} \ast_{\pi_1(S^0)} \{0\}) \ast (\{0\} \ast \{0\})$.
I know it looks ridiculous.
I don't know how to compute this product with amalgamation.
Best Answer
A relatively simple way to describe the fundamental group is as follows:
Consider the eight directions we can go from $(0, 0, 0)$; let $D$ be the set of these directions. Then the fundamental group is the group generated by $D^2$, with the relations that for all $a, b, c \in D$, $(a, b) \cdot (b, c) = (a, c)$.
The idea is that $(a, b)$ represents leaving the origin in the direction $a$, walking around on the sphere a bit, and then coming back to the origin in direction $b$.
But note that your method doesn't work since $S^2 \cup I_x$ is not an open set. You need to use the van Kampen theorem for fundamental groupoids.