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From Wikipedia (about homotopy of spheres):

$\pi_2 (S^1) = 0$: This is because $S^1$ has the real line as its universal cover which is contractible (it has the homotopy type of a point). In addition, because $S^2$ is simply connected, by the lifting criterion, any map from $S^2$ to $S^1$ can be lifted to a map into the real line and the nullhomotopy descends to the downstairs space.

Question 1: Am I right in assuming that I don't need a universal cover? (just any covering space is good enough to apply the lifting criterion)

The "lifting criterion" tells me that if $p: (\tilde{X},a) \to (X,b)$ is a covering map and $f: (Y,c) \to (X,b)$ is any continuous map then $f$ lifts to $\tilde{f}: (Y,c) \to (\tilde{X}, a)$ if and only if $f^\ast \pi_1 (Y) \subset p^\ast \pi_1 (\tilde{X})$.

Hence For $\tilde{X} = \mathbb R, X = S^1, Y = S^2$, $p^\ast \pi_1 (\tilde{X}) = \{0\}$ and hence since $f : S^2 \to S^1$ lifts to $\tilde{f} : S^2 \to \mathbb R$, $f^\ast \pi_1 (Y)$ must be the trivial group, too.

Question 2: But how do I get from there to $\pi_{\color{red}{2}} = 0$? The lifting criterion only tells me something about $\pi_1$.

Thanks for your help.

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Why is suddenly $f$ from the interval to $S^1$? It should go from $S^2$ to $S^1$ (and then $\widetilde f : S^2 \to \mathbb R$). –  PseudoNeo Aug 24 '12 at 14:59
    
@PseudoNeo Sorry, that's a typo. I rewrote the question twice before I posted it. –  Matt N. Aug 24 '12 at 17:39
    
@Matt May be helpful: math.stackexchange.com/questions/184208/… –  user38268 Aug 25 '12 at 8:35

1 Answer 1

up vote 4 down vote accepted

You're right, the claim made by wikipedia is valid for any covering map. And the proof is also valid.

Let $p : \widetilde X \to X$ a covering map.

As you said, the lifting criterion clearly implies that any map $f$ from a simply connected space to $X$ lifts to $\widetilde X$. So, for example, any $f : S^n \to X$ ($n \geq 2$) lifts to $\widetilde f : S^n \to \widetilde X$.

In your case, $X$ is contractible, so there is an homotopy $(\widetilde f_t)$ from $\widetilde f = \widetilde f_0$ to a constant map $\widetilde f_1$. The homotopy $f_t = p \circ \widetilde f_t$ then goes from $f_0 = f$ to a constant map. We have proved that $\pi_n X = 0$ as soon as $\pi_n \widetilde X = 0$.

In fact, one can see along the same lines that every covering map $p : \widetilde X \to X$ induces isomorphisms $p_* : \pi_n \widetilde X \to \pi_n X$ for all $n \geq 2$ (doesn't work for $n=1$, of course: $S^1$ isn't simply connected).

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Cf. also Hatcher, Algebraic Topology, prop 4.1 on p. 342. –  PseudoNeo Aug 24 '12 at 15:10

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