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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.2012-08-05
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    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 $\pi_2(X)\cong\pi_2(\overline{X})\cong\mathbb{Z}^\infty$.2012-08-05
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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.2012-08-05
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    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?2012-08-05
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    @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.2012-08-06
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    @SteveD: I think your wording is overly and unnecessarily harsh. You could give the exact same advice in a much friendlier tone. In particular, it's seriously condescending (not to mention absurd in its vagueness) to tell someone not to ask another question here until they "understand what algebraic topology is all about". What does that even mean??!? I think it's safe to assume that the OP isn't trying to waste anyone's time here; in any case, it's certainly worth giving the benefit of the doubt.2012-08-07
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    @user32240: This question is almost certainly covered in Chapter 1 of Hatcher's book. I think it would do you well to study from there, since he writes clearly and puts a strong emphasis on geometric intuition. At this stage, reading a book would probably be much more beneficial for you than simply running through problems, since books are meant to be comprehensive. Also, I agree that the thing with latex is annoying. It's probably due to the fact that the symbol $\wedge$ is used for the wedge product of differential forms. The symbol $\vee$ is \vee.2012-08-07
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    @AaronMazel-Gee: I have not read Hatcher for two years so my mind is really muddled. When the new semester begins I guess I have to follow his advice to go over it again( I do not have the book at hand). May I ask why the homotopy extension property works for spheres? I did not find the proof on the wiki article.2012-08-07
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    @user32240: Hatcher's book is available free -- you can download it on his website. As for your question: The homotopy extension property is a property of fibrations, of which covering spaces are a particularly nice special case. The point is that it holds for all spaces (or just all CW-complexes, depending on if you're talking about Hurewicz or Serre fibrations resp.), so certainly it's going to hold for $S^2$ either way. I agree with SteveD that you should familiarize yourself with Hatcher chapters 1-3 before worrying about fibrations, though.2012-08-07
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    @AaronMazel-Gee: I had a homotopy rhetoric course two years ago, so I used to work with $\pi_{3}S^{2}$, loop spaces. But now my memory is totally falling apart. Thank you for the advice.2012-08-07
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    @user32240: No problem. Everyone forgets things that they haven't used in a while. Luckily, things are usually easier the second time around!2012-08-07

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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.