Homology of the Klein Bottle I know that in general, $H_{n}(X)$ counts the number of $n$-cycles that are not $n$-boundaries of a simplicial complex $X$. So for the sphere, $H_{0}(X) \cong \mathbb{Z}$ since it is connected. Also $H_{n}(X) = 0$ for all $n>0$ (e.g. all $1$-cycles are $1$-boundaries, etc..). How do you use this geometric interpretation to deduce that $H_{1}(X) = \mathbb{Z} \times \mathbb{Z}_2$ where $X$ is the Klein bottle? This doesn't seem to correspond to the Betti number. 

The second Z-homology of the Klein bottle is zero because it is a non-orientable surface. So the only homology group to compute is the first. 
The fundamental group of the Klein bottle is isomorphic to the group of isometries of the plane generated by the standard lattice (all pairs (x,y) where x and y are both integers) and the map (x,y) -> (x + 1/2, -y).
One computes that the commutator subgroup is generated by the lattice points (0,2m).
In the quotient group, (0,m) generates a subgroup of order 2 and (x + 1/2, -y) generates a copy of Z.
