Computing the homology of a triangle with the edges identified in cyclic order Can someone help me with computing the simplicial homology of the following space, and also clarify why it is a delta complex in the first place:

(The picture means that the edges are identified in the given orientation, also please note that the triangle is a solid triangle).
Thanks in advance!
 A: I think the best way is to subdivide the triangle like that:

The arrows in red aren't identifications: I included them to make the Delta-complex structure clear. Then you get a Delta-complex with three 2-cells, four 1-cells and one 0-cell. Computing the homology of this space is a simple matter of linear algebra, which I'll let you do.
(By the way, if you know cellular homology, the answer would be immediate)

Why is the original not a $\Delta$-complex? Number the vertices of the standard $2$-simplex $\Delta^2 = [0,1,2]$ starting from the bottom left and counter-clockwise. In the quotient space where you identify all the $a$, the edge $[0,2]$ will be the edge $a$ but going in the reverse direction, that is to say $-a$. But $-a$ is not a 1-cell; $a$ is. In the definition of a $\Delta$-complex, all the faces of a cell must again be a cell, and the ordering of the vertices matter. In this case the face $[0,2]$ (the $d_1$ face if it makes sense) will not be one of the $1$-cell, it will be the opposite cell.
