The quotient is obtained from its image $Z\cong S^1$ by attaching the interior of $D^2$. The boundary map winds $n$ times around $Z$, so this is the <cyclic Moore space in dimension one> $M(\mathbb Z/n,1)$. Its <cellular chain complex> is
$$
0\longrightarrow C_2\cong\mathbb Z
\xrightarrow{\ n\ }C_1\cong\mathbb Z
\xrightarrow{\ 0\ }C_0\cong\mathbb Z
\longrightarrow0.
$$
Therefore
$$
\boxed{H_i(Y;\mathbb Z)\cong
\begin{cases}
\mathbb Z,&i=0,\\
\mathbb Z/n,&i=1,\\
0,&i\geq2.
\end{cases}}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 114 / 1 / a / Solution

{title2=$M(\mathbb Z/n,1)$}

Attaching a two-cell to $S^1$ by a map of degree $n$ gives the Moore space $M(\mathbb Z/n,1)$. Its positive-dimensional <cellular chain complex> is
$$
0\longrightarrow\mathbb Z\xrightarrow{\ n\ }\mathbb Z\longrightarrow0,
$$
so $H_1\cong\mathbb Z/n$ and all its other reduced integral homology groups vanish.

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 Incoming links: Cyclic Moore space in dimension one