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Let $P$ be a finite polyhedron and $N$ be a normal subgroup of $G=\pi_1 (P)$. It is known that there exists a covering space $(\tilde{P},p)$ so that $p_* \pi_1 (\tilde{P})=N$. It follows that for the finite polyhedron $\tilde{P}$ (which is related to $P$), we have $\pi_1 (\tilde{P})\cong N$.

My question is that:

Is there any finite polyhedron $Q$ (connected to $P$) so that $\pi_1 (Q)\cong \frac{G}{N}$?

Thanks in advance.

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    $\begingroup$ Seems implausible, consider the case where $P$ is a graph. The quotient group can be an arbitrary finitely generated group, and while that is the fundamental group of a two-complex, it seems unlikely that this complex would have any intimate connection with the graph you started with. $\endgroup$
    – Igor Rivin
    Commented Oct 19, 2017 at 4:17
  • $\begingroup$ Note that $\tilde{P}$ is only finite when $N$ has finite index in $G$. $\endgroup$
    – HJRW
    Commented Oct 19, 2017 at 12:05
  • $\begingroup$ Let $\{r_i\}$ normally generate $N$. Represent (the conjugacy class of) each $r_i$ by a loop in the 1-skeleton of $P$. We can now define $Q$ by gluing a disc to $P$ for each $r_i$, using the corresponding loop as the attaching map. This defines a cell complex that contains $P$. Of course, it can be taken to be finite if and only if $N$ is finitely generated as a normal subgroup. It's unclear to me whether it's "connected to $P$" in the way that you want. (@IgorRivin thinks not, but I prefer to leave it up to the OP to decide.) $\endgroup$
    – HJRW
    Commented Oct 19, 2017 at 12:11
  • $\begingroup$ Note that if, as you seem to imply in your question, you are actually interested in the case that $N$ has finite index in $G$, then there is a canonical finite choice: take the mapping cylinder of the covering map $\tilde{P}\to P$, and then cone off the end isomorphic to $\tilde{P}$. $\endgroup$
    – HJRW
    Commented Oct 19, 2017 at 12:12
  • $\begingroup$ @IgorRivin Thank you for the comment. You are right. This may happen. I check that. $\endgroup$
    – MHenry
    Commented Oct 19, 2017 at 14:01

1 Answer 1

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For a finite polyhedron $P$ and finite-index normal subgroup $N$ of $G=\pi_1P$, there is a canonical finite polyhedron $Q$ with $\pi_1Q\cong G/N$ constructed as follows. Let $\tilde{P}\stackrel{p}{\to} P$ be the covering map corresponding to $N$, as in the question.

We now construct the polyhedron $Q$ as follows:

$Q= ((\tilde{P}\times [0,1]) \sqcup P) / \sim$

where $(\tilde{x},0)\sim (\tilde{y},0)$ and $(\tilde{x},1)\sim p(\tilde{x})$, for all $\tilde{x},\tilde{y}\in\tilde{P}$.

That is, $Q$ is constructed from the mapping cylinder of the covering map $p$ by crushing the canonical copy of $\tilde{P}$ to a point. Alternatively, as I said in comments, one can think of this as obtained by gluing the cone on $\tilde{P}$ to the mapping cylinder of $p$.

The Seifert-van Kampen theorem then tells us that $\pi_1Q\cong \pi_1P/p_*\pi_1\tilde{P}\cong G/N$, as required.

Clearly, this construction can be performed for any subgroup $N$ of $G$, but only gives a finite polyhedron in the case when $N$ has finite index. By making a choice, one may construct a suitable (non-canonical) polyhedron whenever $G/N$ is finitely presentable.

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  • $\begingroup$ Thank you so much for your great answer. This is exactly what I want and expect. That's perfect. Thank you for taking the trouble to help me. I do appreciate you. $\endgroup$
    – MHenry
    Commented Oct 19, 2017 at 15:38

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