= Solution
Use the usual cell structure on the <torus>: one vertex, two one-cells $a,b$, and one two-cell attached by the word $aba^{-1}b^{-1}$. 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 $bb^{-1}$ 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> are
$$
\boxed{H_0=\mathbb Z,\quad H_1=\mathbb Z,\quad H_2=\mathbb Z,\quad H_j=0\ (j\geq3).}
$$
The <pinched torus> is connected, so reduced $H_0$ is zero.
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