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$X_{Ab}$ is the Quillenization of a path-connected space $X$ if $X_{Ab}$ has abelian fundamental group, and there exists a continuous map $X\rightarrow X_{Ab}$ inducing an isomorphism $H_n(X)\rightarrow H_n(X_{Ab})$ for all $n$.

Let $\pi=\pi_1(X)$. The claim that I don't understand is that $X_{Ab}$ exists if and only if $[\pi,\pi]=[\pi,[\pi, \pi]]$.

It seemed at first to me that the proper approach to seeing this would use the result that $f:S^1\rightarrow X$ is in the kernel of the Hurewicz map iff it can be extended to a map of a "disk with handles", which easy to see since generators of $[\pi,\pi]$ form 1-skeletons of closed oriented surfaces. Now, $[\pi,[\pi,\pi]]\subseteq[\pi,\pi]$ in general, and classification of surfaces gives that elements of $[\pi,[\pi,\pi]]$ are 1-skeletons of those oriented surfaces with an odd number of handles, if I've done my counting correctly (which is a tenuous claim at best...)

Trouble is, I don't see how this implies the statement at all. If $f:X\rightarrow X_{Ab}$ is the obvious map which collapses loops in the 1-skeleton of $X$ homotopic to elements of $[\pi,\pi]$, then it clearly induces an isomorphism on homology for $n>2$, and for $n=2$, it doesn't affect $\ker\partial_2$, since the only really troublesome case is $H_1$. Why shouldn't this $f$ induce an isomorphism there, too? It's just sending things to 0 which would get sent to 0 anyway by the Hurewicz map.

I'm not looking for an answer, but if someone could point out at what point I'm clearly thinking about this incorrectly, I'd greatly appreciate it.

gmoss
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    It does affect the kernel of $\partial_2$. Say you have a $2$-chain whose boundary is a loop. If you collapse some loops in the $1$-skeleton, then your chain might suddenly become a new cycle. – Tyler Lawson Feb 19 '13 at 18:00
  • Right... I'll chalk that embarrassing mistake up to writing that after my bedtime. It's still unclear to me how the algebraic condition would affect that, but I'll give it more thought. – gmoss Feb 19 '13 at 19:58

1 Answers1

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Here's what I've come up with so far:

If the condition as stated is incorrect, and is actually $[\pi,\pi]=[[\pi,\pi],[\pi,\pi]]$:

Then first you can prove it fairly easily in the case that $\pi=[\pi,\pi]$, by explicitly constructing the map that glues 2-cells and 3-cells into X so as to preserve $H_2$. Then you take the covering space of $X$, $Y$ which has $\pi_1=[\pi,\pi]$, and, with the corrected condition above, take $Y_{Ab}$ by the previous sentence, and then take the pushout $X\oplus_{Y}Y_{Ab}$, and show that it fits the required description.

Still not clear on whether this condition, which is stronger than the stated one, is correct, and if it is, how to prove the reverse implication.

gmoss
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