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From Harvard qualification exam, 1990. Consider the space $X=\mathbb{S}^{1}\wedge \mathbb{S}^{2}$, alternatively viewing it as a sphere with north and south poles connected. I was asked to:

1): the relationship between $\pi_{2}X$ and $\pi_{2}\overline{X}$, where $\overline{X}$ is $X$'s universal cover.

2): Calculate $\pi_{2}(X)$.

I think $\overline{X}$ is just $\mathbb{R}^{2}\wedge \mathbb{R}^{1}$, and its second homotopy group should be $0$. But I do not see any nontrivial relationship between $\pi_{2}X$ and $\pi_{2}\overline{X}$, for the deck transformation argument does not extend to spheres(as opposed to loops). For the second question, my guess is $\pi_{2}(X)=\mathbb{Z}$; again I need a proof. Does the relationship $$\pi_{2}(X\wedge Y)=\pi_{2}(X)\oplus \pi_{2}(Y)$$ hold as fundamental groups?

I do not see a nontrivial fibration from $\overline{X}\rightarrow X$ that can make me use the long exact sequence of homotopy groups. So I ask in here.

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OK a few things: I think you mean wedge product $S^1\vee S^2$ in your statement, not smash product: the smash product $S^1\wedge S^2$ is just $S^3$. Second, that is not the right universal cover. To get the right one, imagine $\mathbb{R}$ covering $S^1$; each integral point goes to your base point. Since you have an $S^2$ attached to your base point in $X$, you should have one in the universal cover as well. So $\overline{X}$ is just a copy of $\mathbb{R}$ with an $S^2$ attached at every integral point. –  user641 Aug 5 '12 at 15:52
    
Furthermore, every element of $\pi_2(X)$ is represented by a map $f:\ S^2\rightarrow X$, and such a map lifts to $\overline{f}:\ S^2\rightarrow\overline{X}$, by the homotopy lifting property. So $\pi_2(X)\cong \pi_2(\overline{X})$. To actually calculate $\pi_2(\overline{X})$, imagine contracting that copy of $\mathbb{R}$ to a point: you would get an infinte wedge of $S^2$. So $\pi_2(X)\cong\pi_2(\overline{X})\cong\mathbb{Z}^\infty$. –  user641 Aug 5 '12 at 15:55
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Perhaps it would be a good idea to go over some algebraic topology essentials. A lot of things you said in this question are wrong, or at least inaccurate. You do mention fibrations, which is overkill, and miss the fact that a covering map is already a fibration itself. –  user641 Aug 5 '12 at 16:13
    
I mean a wedge product, not a smash product. But latex $\wedge$ gives me the symbol of the smash product, so obviously something is wrong. I need to look up homotopy lifting property - I have no idea that covering maps are fibration - you mean the number of sheets as a discrete set? –  Bombyx mori Aug 5 '12 at 21:50
    
@SteveD: I have not studied covering spaces for a few years, so I need to pick up the basics by studying the problems. I do not quite get your last line - I am not a serious practitioner in at this stage, as I hardly use covering maps in my research. –  Bombyx mori Aug 6 '12 at 15:15
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up vote 3 down vote accepted

I think your mistake is the following: the universal cover of $\mathbb S^2$ is not $\mathbb R^2$. It is just $\mathbb S^2$ because $\mathbb S^2$ is already simply connected.

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