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.

  • $\begingroup$ did you already do the analog case of the circle $S^1$ with the two "intervals" $[-1,1]$ and $[-i,i]$? $\endgroup$
    – janmarqz
    Jul 16, 2021 at 18:08
  • $\begingroup$ This space is homotopy equivalent to $S^2 \vee S^1 \vee S^1 \vee S^1$. Can you see why? $\endgroup$ Jul 16, 2021 at 18:16

1 Answer 1


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.


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