Need help to solve this problem (which is in "Algebraic Topology" by Hatcher):

Let $X$ be the quotient space of $S^2$ obtained by identifying the north and south poles to a single point. Put a cell complex structure on $X$ and use it to compute $\Pi_1(X)$.

In this problem, $D^3$ is the unit disk in $\mathbb{R}^3$ centered in the origin; $S^2 \subseteq \mathbb{R}^3$ the sphere; and ${\partial}(D^3)$ is the boundary of $D^3$.

I tried to think like this: if I'm identifying the north and south poles, then my space is something like $X = S^2/\{ \mathrm{North} \sim \mathrm{South} \}$, so it's like two balloons which has a single point in common. So, if I put a cell complex structure on $X$, the natural try is consider $D^3/{\partial}(D^3) \vee D^3/{\partial}(D^3)$. But $\partial(D^3) = S^2$, so my space is $X = S^2 \vee S^2$.

Using this, the fundamental group of $X$ should be

$$ \Pi_1(X) = \Pi_1(S^2 \vee S^2) = \{ 0 \} $$

because $\Pi_1(S^2) = \{ 0 \}$. So, $X$ is simply connected.

Is my reasoning right? Thanks for your help!

  • $\begingroup$ It is not just $S^2 \vee S^2$. You can imagine taking a line segment that connects the north pole to the south pole. If you take this line to be on the outside of the sphere, you can shrink the space down to a wedge, just not of two $S^2$'s. $\endgroup$
    – J126
    Sep 14 '16 at 1:49
  • $\begingroup$ So, thinking on what you suggested, I noticed that the space $X$ is formed by two spheres, $S^2$ and a segment of line, which separates the spheres. With this in mind, there are 2 kind of loops: a class $[f]$, which is in one sphere, and other $[g]$ in the other sphere. So, $\Pi_1(X) = \mathbb{Z}_2$. $\endgroup$
    – Alexei0709
    Sep 15 '16 at 16:48
  • $\begingroup$ No. It is one $S^2$ with a line attached. $\endgroup$
    – J126
    Sep 15 '16 at 19:17

I can't post this as a comment, but this question has some discussion of a similar space and pictures of the space in question that might help: Are these two spaces homotopy equivalent? .


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