# Fundamental group of a sphere with straight line joining the north and south poles

This problem comes from the book Topology and Geometry of Bredon :

Let $$X$$ be the union of the unit sphere in 3-space with the straight line segment from the north pole to the south pole. Find $$\pi_1(X)$$.

I am reading Atcher's book at the same time. He gives a CW-complex structure to this space, and since the line is contractible, he identifies this space to $$S^2 \vee S^1$$ from which we can compute the fundamental group using Seifert-van Kampen's theorem and we get $$\begin{equation*} \pi_1(X) \approx \mathbb{Z}. \end{equation*}$$

Nevertheless, one should be able to use Seifert-van Kampen's theorem directly to this space without relying on some CW-complex structure as it is not yet introduced in the book of Bredon at this stage. I have tried the following :

Consider the open cover $$\{U, V\}$$ where $$U$$ is a thickened neighborhood of the union of the north hemisphere and the middle-top part of the straight line, and where $$V$$ is a thickened neighborhood of the union of the south hemisphere and the middle-bottom part of the straight line. But, the way I understand it, they are both contractible (they contract to the north and south pole respectively), and thus the fundamental group vanishes by Seifert-van Kampen's theorem.

Where is the error in this argument?

• The intersection of $U$ and $V$ is not path connected! You have two connected components, namely some intersection along the equator of the sphere and the intersection of the two parts of the line. Commented Jan 24, 2021 at 9:15
• Thank you @guidoar for pointing it out! I forgot about this hypothesis in the theorem. Commented Jan 24, 2021 at 9:18
• I tend to forget that one too :) I am currently playing around with different open sets $U$ and $V$, to see if your idea can be fixed, but I keep running into the same issue. Commented Jan 24, 2021 at 9:19

## 1 Answer

Consider $$U$$ and $$V$$ to be semispheres, i.e. $$U$$ goes from the north pole to a bit past the equator, and $$V$$ from a bit above the equator to the south pole. But don't imagine the line to be perpendicular to the equator plane but rather contained in it. Or if you will, construct $$U$$ and $$V$$ relative to Greenwich's meridian :)

If you stare at $$U$$ and $$V$$ for a while, they are (homotopic to) disks with a cord attached to two points of the boundary. Collapsing the disks, one gets moreover a homotopy equivalence with $$S^1$$ in both cases.

The intersection is now a band along the equation together with the line. Via a deformation retraction, this is homotopy equivalent to just the equator - i.e. $$S^1$$ - and the line, which is like a figure $$8$$ (collapse the line).

Hence by van Kampen we have $$\pi_1(X) = \pi_1(U) \ast \pi_1(V) / \langle \langle \iota_1(\omega)\iota_2(\omega)^{-1} : \omega \in \pi_1(U \cap V)\rangle\rangle$$, and $$\pi_1(U) = \pi_1(V) = \Bbb Z$$.

The intersection $$U \cap V$$, as we have said, is homotopy equivalent to a figure $$8$$, whose fundamental group is the free group on two generators, one for each loop. Chasing the homotopies, we obtain that the generators for $$\pi_1(U \cap V)$$ correspond to going through the line and then to one half of each semicircle of the equator. If $$a$$ and $$b$$ are the generators of $$\pi_1(U)$$ and $$\pi_1(V)$$, then these loops correspond to $$a$$ and $$b$$ so the relation introduced is $$a = b$$.

Finally, this gives $$\pi_1(X) = \langle a,b : a = b\rangle \simeq \Bbb Z$$, as desired. Maybe a different choice of orientations changes the presentation a little bit (e.g. the relation could be $$ab = 1$$, which still yields the correct group).