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Let $X$ be a discrete subset of a $K(\pi,1)$. Is the quotient space $K(\pi,1)/X$ still a $K(\pi',1)$? If so, what is $\pi'$?

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1 Answer 1

The quotient of a pathconnected space $A$ by an $n$-point subspace is homotopy equivalent to $A\vee \bigvee\limits_{n-1} S^1$. And $K(\pi,1)\vee S^1=K(\pi*\mathbb Z,1)$. So the answer is, yes, for $\pi'=\pi*F_{n-1}$.

Upd. To make the answer more... self-contained, let me also sketch the proof of the first claim for $n=2$. The crucial lemma: if $X$ is a CW-complex, $Y\subset X$ -- a contractible subcomplex, then $X/Y\cong X$ (the proof can be found in any AT textbook -- for example, see Proposition 0.17 of Hatcher's book).

Now for any 2-point subset $\{a,b\}$ of $A$ take $X=A\cup[0,1]/(0\sim a,1\sim b)$ and apply the lemma for $Y_1=[0,1]$ and for $Y_2$ equal to path $\gamma$ from $a$ to $b$ in $X$ (actually, one has to be slightly more careful here -- because sometimes such path is not a subcomplex etc, but let's ignore this for now): $A/\{a,b\}=X/Y_1\cong X\cong X/Y_2=(A/\gamma)\vee S^1\cong A\vee S^1$.

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You're welcome! –  Alexander Thumm Dec 4 '11 at 19:01
    
Is the quotient map given by $1_A \vee *:A\to A\vee \bigvee_{n-1} S^1$? I don't quite understand how you know the quotient space to be $A\vee \bigvee_{n-1} S^1$ –  user20619 Dec 5 '11 at 15:42
    
So is the homotopy equivalence from $A/\{a,b\}$ to $A\vee_a S^1$ induced from the obvious map $A\to A\vee_a S^1$? –  user20619 Dec 6 '11 at 13:08
    
@user20619 See the update. –  Grigory M Dec 9 '11 at 12:57

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