Two-complexes in Thurston's "Conway's Tiling Groups" I'm trying to read "Conway's Tiling Groups" by W.M. Thurston.  Frankly, I'm more interested in turning some of the algorithms into computer programs, just for my own amusement, than in understanding the math in complete detail, but I want to understand what's going on.
Thurston starts by defining the (Cayley) graph $\Gamma(G)$ of a finitely- generated group $G$.

If $G$ is a group, then its graph $\Gamma(G)$ with respect to generators $g_1,g_2,\dots,g_n$ is a directed graph whose vertices are the elements of the group.  For each vertex $v\in\Gamma(G),$ there will be $n$ outgoing edges, labeled by the generators, and $n$ incoming edges: the edge labeled $g_i$ connects $v$ to $vg_i.$ 
It is convenient to make a slight modification of this picture when a generator $g_i$ has order $2.$  In that case, instead of drawing an arrow from $v$ to $vg_i$ and another arrow from $vg_i$ back to $v,$ we draw a single undirected edge labeled $g_i.$ 

So far, so good.  My problem comes when he extends this graph to a $2-$complex.

Whenever $R$ is a relator for the group, that is, a word in the generators which represents $1$, then if you start from $v\in\Gamma(G)$ and trace out $R,$ you get back to $v$ again.  If $G$ has presentation $$
G=\langle g_1,g_2,\dots,g_n|R_1=1,R_2=1,\dots,R_k=1\rangle
$$ the graph $\Gamma(G)$ extends to a $2-$complex $\Gamma^2(G)$: sew $k$ disks at each vertex of [sic] $v\in \Gamma(G),$ one for each relator $R_i$ so that its boundary traces out the word $R_i.$  An exception is made here for relations of the form $g_i^2=1$ since this relation is already incorporated in making $g_i$ an undirected edge.

Thurston goes on to claim that $\Gamma^2(G)$ is simply connected.  Given a loop, the sequences of edges in the path is a word in the generators that represents the identity. He says,

A proof that this word represents the identity by making substitutions using the relations $R_i$ can be translated geometrically into a homotopy of the path in $\Gamma^2(G).$  

I don't understand at all how the $k$ disk are supposed to be sewn on, though Thurston seems to feel that this is obvious.  I'm thinking of the disk $R_i$ as a regular $m-$gon, where the are $m$ symbols in $R_i$ and the edges of the $m-$gon are labeled in sequence with the symbols of $R_i.$  Are these edges all supposed to be sewn to edges emanating from $v$?  That would seem to require self-intersection, at least in some cases, but later on, Thurston seems to imply that $\Gamma^2(G)$ is embedded in three-space.  (That may be a false impression.  I've only gone through the paper once quickly to determine that it interested me, and now I'm trying to understand it-- and I'm stuck on the first page!) 
I'm also wondering exactly what a $2-$complex is.  I'm thinking it's something like a simplicial complex of dimension $2$, but there you never get anything like self-intersection do you?  (It's been many years since I took algebraic topology, and what little I think I remember is probably wrong.)  
 A: First of all, the difference between a simplicial complex and a cell complex as used in this problem is simply that for a simplicial complex you identify sides of triangles, aka 3-sided polygons, whereas for these more general cell complexes you identify sides of polygons without the restriction to having 3 sides. 
In going from $\Gamma(G)$ to $\Gamma^2(G)$, start from a relator $R$ and a $v \in \Gamma(G)$. The relator $R$ is a certain word in the generating set:
$$R = g_{i_1} g_{i_2} ... g_{i_m}
$$
Associated to the choice of $R$ and $v$ we will sew in a disc $D$ (perhaps a more formal notation would be $D=D(R,v)$, i.e. a different disc for each ordered pair $(R,v)$). 
To do this, first subdivide the boundary of $D$ into $m$ arcs, or simply think of $D$ as a polygon with $m$-sides. 
Now label the sides of that $D$ in order as $g_{i_1}$, $g_{i_2}$, ..., $g_{i_m}$. 
The edges of $D$ are not all sewn on at $v$. Instead, here's what happens.
Starting at vertex $v=v_0$, glue the $g_{i_1}$ side of $D$ to the $g_{i_1}$ edge of $\Gamma(G)$ having initial endpoint $v_0$, and let $v_1$ be the opposite endpoint of that edge.
Next, starting at $v_1$, glue the $g_{i_2}$ side of $D$ to the $g_{i_2}$ edge of $\Gamma(G)$ having initial endpoint $v_1$, and let $v_2$ be the opposite endpoint of that edge.
.
.
.
Continue inductively
.
.
.
In the last step, starting at $v_{m-1}$, will glue the $g_{i_m}$ side of $D$ to the $g_{i_m}$ edge of $\Gamma(G)$ having initial endpoint $v_{m-1}$, and let $v_m$ be the opposite endpoint of that edge. 
And that final vertex $v_m$ is guaranteed to be equal to the original starting vertex $v=v_0$. 
The intuitive reason this complex $\Gamma^2(G)$ is simply connected is because of the construction: the defining relators have all been killed off by sewing in discs. If one wanted to verify formally that this is true, one would have to apply the general Van Kampen theorem.
Finally, there is no implication that $\Gamma^2(G)$ embeds in 3-space. The construction of $\Gamma^2(G)$ is being carried out on an abstract level, using the quotient topology.
