I am trying to compute the fundamental group of $S^2 \cup (0,0,t), t\in\mathbb{R}$. I can come up with 2 different ways of employing deformation retracts and homeomorphic equivalents to use Van Kampen's theorem to compute the fundamental group of this space.

1. enter image description here 2. enter image description here

By (1) I will get a fundamental group given by the fundamental group of $S^2\vee S^1$, i.e $\mathbb{Z}$, whereas by (2) I should get the trivial group since the hemispheres with poles would retract to a single point and the intersection (which is a line) would also have trivial loops.

I can't seem to find out what is going wrong here, both seem reasonable lines of argumentation to me.

Any and all help is appreciated.


1 Answer 1


Your first method is totally correct.

However there are problems with your application of Van Kampen. Firstly you have to choose $U$ and $V$ such that $U\cup V=X$ (the space) and $U\cap V$ is path connected.

Here you have chosen $U$ to be the lower hemisphere along with the axis say $\{(0,0,t):-1\leq t <0.5\}$ (say). But for $U$ to be open, you need to have $U$ to be the open lower hemisphere ( that is the lower hemisphere minus the equatorial circle) along with the axis segment $\{(0,0,t):-1\leq t <0.5\}$ . And similarly for $V$. But in that case, you do not end up with the whole space as the union as you miss out the equatorial circle. In particular, your choice of $U$ is not open (as the points along the equatorial circle would fail to be interior points) .

I'll use some geographical language here so please refer to the picture if you don't already know these enter image description here

Now to remedy this, you should take $U$ to be the lower hemisphere along with some portion of the "torrid zone" . You can take $U$ to be the lower hemisphere and add to it the portion upto the Tropic of cancer minus the circle at the tropic of cancer (to make it open) and also add the segment $\{(0,0,t):-1\leq t <0.5\}$

Similarly take $V$ to be the upper hemisphere along with the portion upto the tropic of capricorn minus the circle at the tropic of capricorn and also add the segment $\{(0,0,t):-0.5< t \leq 1\}$ .

Then the intersection would be like a play top which would be homeomorphic to the cylinder along with it's axis . i.e. $\{(x,y,z):x^{2}+y^{2}=1\,,0\leq z\leq 1\}\bigcup\{(0,0,t):0\leq t\leq 1\}$ . This would not have trivial fundamental group.

The point is that applying Van Kampen in this case is not as easy unless you can come up with some clever CW complex structure . Of the top of my head, a CW complex structure is perhaps like this: - first picture reveals that you have one 0-cell, two 1 cells and one 2-cell along one of the one cell but that would again just lead you to your first method.

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    $\begingroup$ Thanks, this clears it up! $\endgroup$ Oct 5, 2022 at 15:33

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