Triangulate the triangle with edges identified

Consider the closed triangle (simplex) $\Delta = [(0,1), (0,0) , (1,0)] \subset \mathbb{R}^2$, ie, $\Delta$ is the convex hull of the points $v_1 = (0,1)$, $v_0 = (0,0)$ and $v_2 = (1,0)$ in $\mathbb{R}^2$, with the usual (euclidean) induced topology from $\mathbb{R}^2$. Identify the points to form the quotient topology like this: all the points of the interior of $\Delta$ are identified just with themselves, ie, form the classes $\{p\}$ for $p \in int \Delta$. On the boundary of $\Delta$, identify sets of 3 points starting on the vertices and going along each edge counterclockwise, ie, form the classes $\{ (0,1-t) , (t,0) , (1-t,t) \}$, for all $t \in I = [0,1]$, ie, for $0 \leqslant t \leqslant 1$.

The quotient space formed is $\Delta$, the triangle with edges identified.

The questions:

(1) There exists a triangulation for $\Delta$? Ie, there exists a simplicial complex homeomorphic to $\Delta$?

(2) How to compute the homology groups of $\Delta$? Here I can use CW-complexes, but I don't know if I performed this correctly.

My calculations gave the same homology groups of sphere S², but my intuition tells me that $H_1 (\Delta)$ should be $\mathbb{Z} / 3 \mathbb{Z}$. – Gustavo Jun 18 '14 at 21:41
You have one $0$-cell, one $1$-cell, and one $2$-cell, so you have the sequence $$0\to\mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}\to 0.$$ The first and last maps are (obviously) zero. So is the third, because the edge has both endpoints on the same vertex. The third map is $\times 3$, because the edge of the triangle goes three times along the edge, in the same direction. – Jessica B Jun 19 '14 at 19:40