Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 1 / 21H / Solution

Use the usual cell structure on the torus: one vertex, two one-cells $a,b$, and one two-cell attached by the word $ab{a}^{-1}{b}^{-1}$. Take the meridian being collapsed as $a$. The quotient has one vertex, the surviving circle $b$, and a two-cell whose attaching word reduces to $b{b}^{-1}$ and is null-homotopic. Attaching a two-cell to a circle by a null-homotopic loop gives a space homotopy equivalent to $S^1\vee S^2$.

Alternatively its cellular chain groups are $C_2=C_1=C_0=\mathbb{Z}$, and both boundary maps vanish: the two-cell traverses $b$ once in each direction, and the one-cell's endpoints coincide. Thus the integral homology groups areThe pinched torus is connected, so reduced $H_0$ is zero.