Computing the homology groups of a given surface Let $\triangle^2=\{(x,y)\in\mathbb{R}^2\mid 0\le x,y\wedge x+y\le1\}$ (that is, a right triangle). Define the equivalence relation $(t,0) \sim (0,t)\sim (t,1-t)$.
I want to compute the homology groups of $\triangle^2/\sim$.
An attempt at doing so was to define $U=\{(x,y)\in\mathbb{R}^2\mid 0< x,y\wedge x+y<1\}$ and $V=\triangle^2 \setminus \{1/3,1/3\}$.
It is clear that $U\cup V = \triangle^2$ and so Mayer-Vietories could be useful here.
Noting the following facts:


*

*$V$ is a retract deformation of the boundary of the triangle and since all lines are identified it is homeomorphic to $S^1$, and so $H_2(V)=0$, $H_1(V)=\mathbb{Z}$ and $\tilde{H}_0(V)=\mathbb{Z}$.

*$U$ is retractable and so it's positive dimension homology groups vanish, and it's zero dimensional homology group is $\mathbb{Z}$

*$U\cap V$ is again homotopy equivalent to $S_1$


At this point it's really easy to see (using M.V) that $H_n(\triangle^2 / \sim ) = 0$ for $n>2$ and also for $n=0$.
For lower values of $n$, taking the M.V. sequence for reduced homologies and plugging in the values I already know, I get.
$0\to H_2(\triangle^2 / \sim ) \to \mathbb{Z} \to \mathbb{Z}\to H_1(\triangle^2 / \sim )\to 0$.
This is the point where I don't know what to do next, and any help would be appreciated.

Just knowing that sequence is exact is not enough since, for example, $H_2(\Delta^2/\sim) = 0 = H_1(\Delta^2/\sim)$ and $H_2(\Delta^2/\sim) = 0, H_1(\Delta^2/\sim) = \mathbb Z/n\mathbb Z$ both work.
So you need to look at the actual map $H_1(U \cap V) \to H_1(U) \oplus H_1(V) \simeq 0 \oplus H_1(V)$, which is given by the two inclusions.  But $U\cap V$ is a deformation retract of $V$ so that the inclusion $U \cap V \to V$ induces an isomorphism on homology.  Thus the map $H_1(U \cap V) \to H_1(U) \oplus H_1(V)$ is an isomorphism so that $H_2(\Delta^2/\sim) = 0 = H_1(\Delta^2/\sim)$.
