Homotopy group of three spheres

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. • I have never studied algebraic topology seriously, but isn't your shape like a torus pinched at three points (and therefore have a fundamental group of $\mathbb{Z^2}$)? – Nick A. May 16 '18 at 16:55
• That's pretty smart! I think that could work yeah. But I am specifically looking for an answer that uses Van Kampen. – Elie Bergman May 16 '18 at 16:58
• Actually this is incorrect sorry. You cannot 'pinch' a space whilst preserving its homotopy type. Consider the pinched sphere for instance. – Elie Bergman May 16 '18 at 17:03
• The pinch actually eliminated one of the circles of $\mathbb Z^2.$ @NickA. – Thomas Andrews May 16 '18 at 17:10
• @ElieBergman All of you are right! Here is a similar question, which also explains why I am wrong : math.stackexchange.com/questions/2081978/…. I hope this helps :) – Nick A. May 16 '18 at 17:10

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 $\mathbb{Z}$.

• Thanks for addressing my wrong comment! +1 – Nick A. May 16 '18 at 17:18
• Thanks for this answer. I am not familiar with Van Kampen applied to more than two subspaces, but I can see how it might work. Might you comment on why the construction I gave failed? (It gave me trivial homotopy group). – Elie Bergman May 16 '18 at 17:19
• @ElieBergman You're missing the part of the intersection that includes the "back curve", so the intersection is disconnected. – Kyle Miller May 16 '18 at 17:22
• A rule of thumb for the van Kampen theorem: all of the generators must be accounted for between the subspaces. It is unable to create new generators (unless you use fundamental groupoids, like Ronnie Brown has written about). – Kyle Miller May 16 '18 at 17:28

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 $\mathbb{Z}$.

• Can you explain this in a bit more detail with regards to the free product with amalgamation that is involved in Van-Kampen? – Elie Bergman May 16 '18 at 17:55
• Also, the intersection is not the wedge of four circles. It is not even the wedge of three circles! – Elie Bergman May 16 '18 at 18:02
• @ElieBergman The intersection is given by the equators of the three spheres, which are arranged "in a triangle". If you draw it, you'll see that it is a wedge of four circles. I am not here to solve your exercise for you, so I will let you work out the details. The best way to learn maths is by doing maths. – Daniel Robert-Nicoud May 16 '18 at 18:08

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 $\mathbb{Z}$.