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Let $Y$ be a CW-complex and $X$ its universal cover. Could you give me a proof (or a referece) for the following fact:

$X$ is contractible $\Leftrightarrow$ $H_i(X)=0$ $\forall i\geq2$ $\Leftrightarrow$ $\pi_iY=0$ $\forall i\geq2$.

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This is the Hurewicz theorem and the Whitehead theorem. The idea is the first non-trivial homology group of a space is the (abelianization of) the first non-trivial homotopy group. That's Hurewicz. The Whitehead theorem says that if a map between CW-complexes is an iso on all homotopy groups then it's a homotopy equivalence -- in the case of a CW-complex with all homotopy groups trivial, take a 1-point subcomplex as your other CW complex. You can find these in any intro alg-top textbook: Spanier, May, Hatcher, Switzer, Whitehead, Kirk&Davis, Harper, and so on. –  Ryan Budney Sep 3 '11 at 21:06
For the second implication, use the fact that any map from $S^n \rightarrow Y$ for $n\ge 2$ lifts to a map $S^n \rightarrow X$. So if $X$ has all trivial homotopy groups, then this will be nullhomotopic in $X$, hence nullhomotopic in $Y$. –  Dylan Wilson Sep 3 '11 at 22:22
You might also find Peter May's paper "On the dual Whitehead Theorems" of interest. –  Dan Ramras Sep 4 '11 at 15:21
I can't understand why $\pi_iY=0$ $\forall i\geq2$ implies $X$ contractible. From Whitehead it's clear if I have $\forall i\geq 1$. Other thing that I don't understand is why $H_i(X)=0$ $\forall i\geq2$ implies $X$ contractible (or why it implies the condition on the homotopy groups). –  Berry Sep 7 '11 at 1:25

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