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Is there a proof of the Hurewicz theorem $\pi_1(X)^{ab} = H_1(X, \mathbf Z)$ ($X$ a connected topological space) expressing $\pi_1(X)$ as the "Galois" group of $X$, i.e., group of deck transformations of the universal cover?

(as opposed to a proof using the construction of $\pi_1$ as the group of paths up to homotopy).

Thank you.

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I don't understand your question. It is certainly true that $\pi_1(X)$ is the group of deck-transformations, but what does this have to do with Hurewicz theorem? –  J.C. Ottem Feb 23 '11 at 19:17
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How do you want to define $H_1(X)$? If you want to use singular homology then it seems unlikely that you can avoid some kind of paths. You could describe $H^1(X)$ as $[X,S^1]$ or using Cech cochains and then consider pairings $\pi_1(X)^{ab}\otimes H^1(X)\to\mathbb{Z}$, but you would lose information about torsion in $H_1(X)$ that way. –  Neil Strickland Feb 23 '11 at 19:18
    
If one defines $H^1(X,{\mathbb Z})$ as the group of ${\mathbb Z})$ torsors over $X$, then one can see that $H^1(X,{\mathbb Z})$ is the group of homomorphisms from $\pi_1(X)^{mathrm ab}$ to ${\mathbb Z}) by thinking about how deck transformations act on torsors. But that is $H^1$ not $H_1$. –  user10849 Feb 23 '11 at 19:57
    
Sorry -- forgot a dollar sign: If one defines $H^1(X,\mathbb{Z})$ as the group of $\mathrm Z$-torsors over $X$, then one can see that $H^1(X,\mathbb{Z})$ is the group of homomorphisms from $\pi_1(X)^{\mathrm{ab}}$ to $\mathbb Z$ by thinking about how deck transformations act on torsors. But that is $H^1$ not $H_1$. –  user10849 Feb 23 '11 at 20:02
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I would be tempted to define $H_1$ as the abelianisation of $\pi_1$ in the general Tannakian setting, and use Hurewicz (when it holds) to say that this new $H_1$ and the old one (simplicial homology, say) agree. –  David Roberts Feb 23 '11 at 23:31
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1 Answer

For $X$ a $0$-connected nice space (say, a CW-complex), and for any group $G$, there is a natural bijection of the following shape

$$[X,BG]\simeq Hom(\pi_1(X),G)$$

which can be proved roughly as follows (if you like tannakian-like arguments): maps from $X$ to $BG$ correspond to $G$-torsor over $X$, which correspond to maps of topoi from the topos of sheaves over $X$ to the topos of $G$-sets; but, as any $G$-torsor is locally constant, this also corresponds to the maps of topoi from the topos of locally constant sheaves over $X$ to the topos of $G$-sets. As, by Galois theory, the topos of locally constant sheaves over $X$ is canonically equivalent to the topos of $\pi_1(X)$-sets, we conclude from the fact that, given two groups $A$ and $B$, exact and colimit preserving functors from $B$-sets to $A$-sets correspond to homomorphisms of groups from $A$ to $B$.

To be precise, $[X,BG]$ means the set of homotopy classes of maps from $X$ to $BG$, while for $G$ an abelian group, $Hom(\pi_1(X),G)$ means the set of group homomorphisms (for a non abelian $G$'s, we have to quotient a little bit, but we won't care here).

For $A$ an abelian group, we thus get bijections $$H^1(X,A)\simeq [X,BA]\simeq Hom(\pi_1(X)^{ab},A) .$$ By the Yoneda lemma, to prove that the map $\pi_1(X)^{ab}\to H_1(X,\mathbf{Z})$ is bijective, it is sufficient to prove that, for any abelian group $A$, the map $$<\star> \quad Hom(H_1(X,\mathbf{Z}),A)\to Hom(\pi_1(X)^{ab},A)$$ is bijective. But, instead of checking this for all $A$'s, it is sufficient to prove this in the case where $A$ is an injective object in the category of abelian groups (because there are enough injectives). In this case, as $Hom(-,A)$ is an exact functor, we have a bijection $$Hom(H_1(X,\mathbf{Z}),A)\simeq H^1(X,A) .$$ Therefore, for any injective $A$, the map $<\star>$ is bijective.

If you like topoi and pro-groups, you may play the same game and prove this for locally $0$-connected topoi with essentially the same argument.

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