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Let $K(G,1)$ denote the Eilenberg-Maclane spaces with fundamental groups isomorphic to $G$.

Consider the category $\mathbf{K^1_{CW,*}}$ where objects are pointed $K(G,1)$ CW complex, and morphisms are equivalence classes of homotopy rel basepoint pointed continuous maps. Then the fundamental group functor $$ \Pi_1:\mathbf{K^1_{CW,*}}\to \mathbf{Grp} $$ is fully faithful. This is by Theorem 1B.8 in Hatcher. Moreover, this functor is essentially surjective, because for every group $G$, we can construct an element of $K(G,1)$ denoted $BG$ constructed in Hatcher 1B.7. Therefore $\Pi_1$ induces an equivalence of categories assuming axiom of choice.

My questions is:
1. Is the above correct?
2. Is the inverse functor given by the construction that assigns to $G$ the $K(G,1)$ $\Delta$-comple BG, and $\phi:G\to H$ the induced morphism $B\phi:BG\to BH$ given by Hatcher 1B.7?
3. Does it generalize to higher homotopy group construction $\Pi_n$ and higher Eilenberg-Maclane spaces $K(G,n)$?

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  • $\begingroup$ How is choice used ? $\endgroup$ – Rene Schipperus Oct 1 '14 at 2:59
  • $\begingroup$ @ReneSchipperus I was told that fully faithful and essential surjective implies equivalence requires AC. $\endgroup$ – mez Oct 1 '14 at 3:00
  • $\begingroup$ Are t he objects spaces or homotopy equivalence classes of spaces ? $\endgroup$ – Rene Schipperus Oct 1 '14 at 3:01
  • $\begingroup$ @ReneSchipperus I'm thinking just spaces. $\endgroup$ – mez Oct 1 '14 at 3:02
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    $\begingroup$ My intuition tells me that in this case choice is not needed since you have a well defined method of constructing a $K(G,1)$ from $G$. $\endgroup$ – Rene Schipperus Oct 1 '14 at 3:15
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I will work with Kan complexes (i.e. fibrant simplicial sets) instead of CW-complexes. For the purposes of homotopy theory there is no essential difference: the homotopy category of pointed Kan complexes is equivalent to the homotopy category of pointed CW-complexes.

Let $\mathbf{Kan}_*$ be the (ordinary) category of pointed Kan complexes. As is well known, we have a functor $$\pi_n : \mathbf{Kan}_* \to \mathbf{Ab}$$ for every integer $n \ge 2$. Slightly less well known is that we have a functor $$K (-, n) : \mathbf{Ab} \to \mathbf{Kan}_*$$ going in the other direction. The quickest way to define it is to say that $K (A, n)$ is (the underlying pointed Kan complex of) the simplicial abelian group corresponding (under Dold–Kan) to the chain complex $A [-n]$ that has $A$ in degree $n$ and $0$ otherwise. More explicitly, $$K (A, n)_m = \bigoplus_{[m] \twoheadrightarrow [n]} A$$ where the direct sum is indexed over the set of monotone surjections $\{ 0, \ldots, m \} \to \{ 0, \ldots, n \}$, with simplicial operators defined as follows: for any monotone map $\alpha : \{ 0, \ldots, l \} \to \{ 0, \ldots, m \}$, given a monotone surjection $\beta : \{ 0, \ldots, m \} \to \{ 0, \ldots, n \}$ and an element $a$ of $A$, if $\beta \circ \alpha$ is surjective, then $\alpha$ sends $(\beta, a)$ to $(\beta \circ \alpha, a)$; otherwise $\alpha$ sends $(\beta, a)$ to $0$.

The Dold–Kan correspondence says that, for any simplicial abelian group $G$, we have a natural isomorphism $\pi_n (G) \cong H_n (C)$, where $C$ is the chain complex corresponding to $G$. In particular, we have $$\pi_n (K (A, n)) \cong H_n (A [-n]) \cong A$$ naturally in $A$. This shows that $\pi_n : \mathbf{Kan}_* \to \mathbf{Ab}$ is essentially surjective on objects, while $K (-, n) : \mathbf{Ab} \to \mathbf{Kan}_*$ is faithful and conservative. With a bit more work it can be shown that $K (-, n) : \mathbf{Ab} \to \mathbf{Kan}_*$ is also full.

Of course, it should go without saying that $\pi_n : \mathbf{Kan}_* \to \mathbf{Ab}$ factors through the homotopy category to give a functor $\pi_n : \operatorname{Ho} \mathbf{Kan}_* \to \mathbf{Ab}$, which will be a left inverse for the functor $K (-, n) : \mathbf{Ab} \to \operatorname{Ho} \mathbf{Kan}_*$, which is full. Thus $\mathbf{Ab}$ is equivalent to the full subcategory of $\operatorname{Ho} \mathbf{Kan}_*$ spanned by the pointed Kan complexes that are homotopy equivalent to $K (A, n)$ for some abelian group $A$.

The case $n = 1$ requires some adjustments, but you seem to be familiar with that already.

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