Fundamental group of the sand clock

Question

I computed the fundamental group of the following figure with the usual topology:

The figure corresponds to the set:

$X = \{(x,y,z) \in \mathbb{R}_3:x^2+y^2 = z^2,|z| \le 1\} \cup \{(-1,0,z):|z| \le 1\} \cup \{(1,0,z):|z| \le 1\}$

And I obtained that it is the trivial group using Seifert-Van Kampen theorem. However, other people claim that it is the free group with two generators.

My solution

I basically took as open sets one of the lines with a cone plus a mini cone. That way I have to open, path-connected-sets with the hypotheses of Seifert-Van Kampen.

Answer 1

So my friend was right the main problem with my solution was that the sets aren't open:

In the image I have removed the point of intersection between the segment and the upper circumference. The remaining points of the open set in the induced topology should be neighborhood of all of its point, but intuitively neighborhoods near the excluded point should contain it.

Instead, once should apply Seifert-Van Kampen theorem by removing points from one segment and one circumference. Then one should deform the figure into a cone with a loop entering it. One then deduces that the fundamental group must have two generators.

I'm happy with this solution (not so happy with the open set stuff though), but what was important is that I learnt the intuition to compute the fundamental group . Here you can have two loops in the figure, that go from the segment to the cones and back to the segment. This are "essentially" the two only loops that one can have.

Answer 2

Another possible solution is to notice that you can deform the sand clock into a figure eight, which immediately gives the result.

Writing down an explicit (sequence of) deformation(s) is an easy, but rather boring exercise.