Triangulation and Homology group of hexagon with neighbouring edges identified I am having trouble with an exercise regarding a special identification space obtained from a hexagon.

Consider a 6-gon with bordering edges identified along the word w=caabbc. Call the resulting quotient space $X$.
a) Triangulate $X$ and call its simplicial complex $K$.
b) Comnpute the Euler characteristic of $X$. Is $X$ orientable?
c) Perform a surgery along a chosen closed non - separating polygonal curve in $K$. Indicate this curve in the triangulation you have drawn for part a). Call the resulting space $Y$ and compute it's Euler characteristic.
d) Compute the homology $H_{*}(Y)$ of $Y$.

Here is my attempt at a solution:
a) Consider the centre of the hexagon (denoted by $x$). Join every vertex of the hexagon to $x$ via a straight edge. This breaks up the hexagon into 6 triangles.
b) Since neighboring edges are identified it follows that the vertices of the hexagon are all identified into one vertex. Together with the centre $x$ this gives us a single edge and two vertices for the triangulation of $X$.
Hence the Euler characteristic is:
$\chi(K)=2-1=1$
Moreover, since the triangulation consists only of a single edge and to vertices it follows that $X$ is orientable.
But now I have gotten stuck at c). I do not see how to draw a closed polygonal curve that does not separate the space because there is only one edge in the triangulation. Of course I could cheat and chose a curve that consists of only a single vertex, but this doesn't seen right.
 A: Your attempt starts off okay, by inserting the vertex $x$, and then counting the identifications of vertices on the boundary. In b), you were okay saying that there are two vertices in $X$. 
But you should think again about counting the edges. When you broke the hexagon into 6 triangles, after inserting a new vertex $x$ at the center of the hexagon, you failed to mention that you must also insert six edges connecting $x$ to each of the six vertices of the hexagon. On top of that you failed to take into account the identifications of the $6$ original edges on the boundary of the hexagon; for example, its certainlhy not true that all 6 of those edges get identified to a single edge of $X$.
And what's more, you failed to count the triangles of the triangulation at all. Keep in mind, for a 2-dimensional triangulation, the formula for Euler characteristic is
$$\chi(X) = \#\text{vertices of $X$} - \#\text{edges of $X$} + \#\text{triangles of $X$}
$$
You entirely omitted the third term. So you should think about the effect on the hexagon itself from inserting $x$ and inserting the six edges that connect $x$ to the six vertices of the hexagon.
Because of your incorrect counts, your computation of the Euler characteristic and your demonstration of orientability are invalid.
