6

What is the fundamental group of $\mathbb S^2$ attached with a diameter? And what is the fundamental group of a hemisphere attached with a diameter?

I guess for the latter one, we can deformation retract it to a circle by compressing around the diameter. So the fundamental group should be $\mathbb Z$. And for the first one we can combine this result with Van Kampen's theorem, and the answer should also be $\mathbb Z$.

Am I right? And how can I prove the case for hemisphere more rigorously?

  • @janmarqz !! So how can I see it? –  Oct 18 '15 at 19:35
  • sorry, I has a doubt... i'm going to delete my comments... i'm going to think it better – janmarqz Oct 18 '15 at 19:44
  • 1
    conpare this answer http://math.stackexchange.com/questions/1470603/practicing-seifert-van-kampen/1472275#1472275 – Ronnie Brown Oct 18 '15 at 19:56
  • @NajibIdrissi I guess so...... So attach a line inside and outside actually do not change the fundamental group right? And what about hemisphere? Oh, what about do not use this collapsing criterion. Because my professor have not introduced it yet. Does there exist other more elementary proof? –  Oct 18 '15 at 20:36
  • I voted to reopen. It is not obvious that the space of this question and the question linked are homotopically equivalent (they are not even homeomorphic). – Aloizio Macedo Oct 19 '15 at 13:14

1 Answers1

5

You nailed space (b). So I'll talk about space (a), which we will denote by $A$.

Space (a) is connected, arcwise connected and semi-locally simply connected. Hence, it has a universal cover. It is natural to try to find out that universal cover. With some thinking, we get that the universal cover is the "Happy Beads" (for the lack of a good non-pornographic word) depicted below:

enter image description here

where the projection map is quite obvious. Now what follows is simply a mere adaptation of the proof that $\pi_1(S^1)=\mathbb{Z}$, which we outline below:

Fix a point $y$ in the universal cover and consider the map $HB: \pi_1(A, x_0) \rightarrow \mathbb{Z}$, where $x_0$ is the upper point of the sphere (which has a segment going down), which takes a class $[f]$ and sends it to $n$, where $n$ is the number of spheres (counted with "direction") between the initial point of the lift of $f$ (which is $y$) and the endpoint. This is well-defined due to the properties of lifts. $HB$ is obviously surjective. Now, if the $n$ associated with a given class is $0$, then the lifted path is a loop in the universal cover. Hence, homotopically trivial. Therefore, we have that $[f]=p_{\#}[Lift ~f]=p_{\#}0=0$. We then have that $HB$ is an isomorphism.

Aloizio Macedo
  • 34,292