I am trying to wrap my head around the following problem: I have three objects lined up horizontally, a $2$-sphere, a circle, and another $2$-sphere. It is the wedge sum $S^2 \vee S^1 \vee S^2$. I am trying to find the fundamental group of this space as well as the covering spaces. For the fundamental group, I believe that I can use van Kampen in the following manner:

$$\begin{align*} \pi_1(S^2 \vee S^1 \vee S^2) &= \pi_1(S^2) * \pi_1(S^1 \vee S^2) \\ \pi_1(S^2 \vee S^1 \vee S^2) &= 0 * \pi_1(S^1 \vee S^2) \\ \pi_1(S^2 \vee S^1 \vee S^2) &= 0 * (\pi_1(S^1) * \pi_1(S^2)) \\ \pi_1(S^2 \vee S^1 \vee S^2) &= 0 * \mathbb Z * 0 \end{align*}$$

Does this make sense?

I am still trying to work out how to find the covering spaces.

  • 1
    $\begingroup$ Let the space be $W$. Then as you have seen Seifert-van Kampen tells us that $\pi_1 (W)\cong \mathbb{Z}$, and more generally $\pi_1(V\vee U)\cong \pi_1(V)*\pi_1(U)$. But this is also intuitively true: Say I had a big loop in our space $W$. We know the fundamental group of the circle is $\mathbb{Z}$, and since any higher dimensional sphere is simply connected, I can just contract the parts of the loop on the $2$-spheres so that the big loop becomes only a loop on the circle. This gives us a correspondence between loops in the space $W$, and loops in the circle. $\endgroup$ – Eric Naslund Dec 21 '11 at 17:12
  • $\begingroup$ The universal cover will be an infinite line with a string of $S^2$'s attached to it. The deck transformations are generated by motion two steps to the right. Do you see how to get the other covering spaces? $\endgroup$ – Cheerful Parsnip Dec 21 '11 at 17:18
  • $\begingroup$ Thanks a lot for the help; I really appreciate it. I'm sorry, but I am not seeing how to get the other covering spaces. What is the best way to begin deriving them? $\endgroup$ – Chammershield Dec 21 '11 at 17:36

You're correct about the computation of $\pi_1$, though depending on the level a bit more work may need to be shown (i.e., how does Seifert- van Kampen give the results you stated). Also, $0*\mathbb{Z}*0$ is naturally isomorphic to $\mathbb{Z}$, and you may want to mention this.

As far as the covering space aspect, it may help to note that $S^2\vee S^1\vee S^2$ is homeomorphic to $S^1\vee S^2\vee S^2$ and you somehow need to "unravel" the $S^1$. Since the universal cover of $S^1$ is $\mathbb{R}$, it should be no surprise that $\mathbb{R}$ enters the picture somehow when finding the universal covering space.

In fact, you might guess that the universal cover is $\mathbb{R}\vee S^2\vee S^2$, since this space is simply connected. Unfortunately, this isn't correct. The problem is that every $2\pi$ along the $\mathbb{R}$ piece should project to the wedge point which has an $S^2\vee S^2$ attached to it. So, our next guess is that the universal cover is a copy of $\mathbb{R}$ with an $S^2\vee S^2$ attached to each point of the form $2\pi k$ for $k\in\mathbb{Z}$. Now that you have the picture in mind, I'll leave it to you to try to prove this space is the universal cover.

More, in fact, is true: If $X$ is simply connected, then the universal cover of $S^1\vee X$ is $\mathbb{R}$ with an $X$ wedged to each point of the form $2\pi k$. Your proof in the $S^2\vee S^2$ case will likely automatically generalize to this statement.

  • $\begingroup$ I think I understand. And by this same logic, would the regular covers be S2∨S2 attached to points of the form 2pik for k∈Z? $\endgroup$ – Chammershield Dec 21 '11 at 17:58
  • $\begingroup$ I don't understand what you're asking. All of the covers, in this case, would be regular. If you attach $S^2\vee S^2$ to poitns of the form $2\pi k$, you get the universal cover. You have to attach $S^2\vee S^2$ to a circle (some number of times) to get the other covering spaces. $\endgroup$ – Jason DeVito Dec 21 '11 at 18:09
  • $\begingroup$ All right i see now. I apologize for the confusion. Thank you for your help! $\endgroup$ – Chammershield Dec 21 '11 at 18:23

The fundamental group of $S^2\vee S^1\vee S^2$ is $\mathbb{Z}$ as you stated ($S^2\vee S^2$ is simply connected, wedge with a circle gets you $\mathbb{Z}$). For each subgroup $n\mathbb{Z}$ of $\mathbb{Z}$ there is a covering space corresponding to that subgroup. These are "bracelets" with a sphere attached at regular intervals ($S^1$ with spheres attached), with $n\mathbb{Z}$ corresponding to the bracelet with $n$ beads. For $n=0$ the universal cover is an infinite bracelet ($\mathbb{R}$ with spheres attached).

This is assuming that the wedge points are distinct, else the "beads" on the bracelet come in pairs


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