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.



On the realization of a quotient group
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.


