Homotopy group of three spheres

Question

Suppose we take three spheres $X_1 , X_2 , X_3$ and identify the north pole of $X_i$ with the south pole of $X_{i+1}$, where $i$ is taken mod $3$. I.e. We sort of have a 'triangle' of spheres. I want to find the fundamental group using Van Kampen's theorem. I can see intuitively that the fundamental group should be $\mathbb{Z}$ but every time I try to calculate, I keep getting a trivial fundamental group. Any ideas here?

In particular, I am specifically interested in why the following application of Van-Kampen fails.

Answer 1

The hypothesis of the van Kampen theorem (at least in Hatcher) is that each subspace contains the basepoint.

A simpler example for what is going wrong: consider a decomposition of $S^1$ into three arcs, every pair of which intersects along an arc. The generator of $\pi_1(S^1)$ can't exactly be represented as loops in each of these three subspaces...

We can still use the van Kampen theorem, but we need to be careful with our choice of subspaces. For each $X_i$, let $\gamma_i$ be a longitudinal path from the north pole to the south pole. The composition $\gamma_1\gamma_2\gamma_3$ is a closed loop. Fix the basepoint on this loop. Take a small neigborhood $U$ of this loop. Now, the three sets $X_i\cup U$ form a decomposition of the space that satisfies the van Kampen theorem's hypotheses. Since $\pi_1(X_i\cup U)=\mathbb{Z}$, you can get the result you want.

Or, an argument that does not use the van Kampen theorem is to manipulate a CW complex to get a homotopy equivalent space. If you contract $\gamma_1$ and $\gamma_2$ from before, you get a space that is the wedge product of two spheres and a sphere whose north and south poles are identified. A sphere whose north and south poles are identified is homotopy equivalent to the wedge sum of a circle and a sphere. Hence, your space is homotopy equivalent to the wedge sum of three spheres and a circle.

Or, following up on a suggestion of Nick A., a torus with three disks glued in along translates of the same simple closed curve is homotopy equivalent to your space. Gluing in a disk is the same as adding a relation to the fundamental group. Any simple closed curve can be used as one of the generators for the fundamental group of a torus. The disks all kill this particular generator, hence all that is left is $\mathbb{Z}$.

Answer 2

Consider the subdivision given by $$U = \{\text{the upper hemisphere of the three spheres}\}$$ and $$V = \{\text{the lower hemisphere of the three spheres}\}\ .$$ Then $U\simeq S^1\simeq V$, and $U\cap V$ is a wedge of four circles. Then applying Seifert-van Kampen we obtain that the fundamental group is $\mathbb{Z}$.

Answer 3

You can use the more general groupoid version of Seifert-van Kampen, which will tell you that the fundamental group(oid) here is the same as if the spheres had been replaced by intervals (basically since they're simply connected). This makes the answer clear: you get a circle this way, so the fundamental group is $\mathbb{Z}$.