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.
Thanks in advance
Taking the second Barycentric subdivision of the triangle (ie. divide it into six, and then divide each of those into 6), and then quotienting, should give a simplicial complex. You could then use simplicial homology, but there are a lot of triangles. In practice I would calculate as if the single triangle did give a simplicial complex.