On the realization of a quotient group 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.
 A: 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.
