Let $T$ be the <surface with boundary> obtained by deleting the interiors of the $n$ disks. By <excision>,
$$
H_i(S,T)\cong\bigoplus_{j=1}^nH_i(D^2,S^1),
$$
so $H_2(S,T)\cong\mathbb Z^n$ and the other relative groups vanish. In the <long exact sequence in relative homology>, the map
$$
H_2(S)\cong\mathbb Z\longrightarrow H_2(S,T)\cong\mathbb Z^n
$$
sends an <orientation class> to $(1,\ldots,1)$, up to an overall choice of sign. It is therefore injective and has <cokernel> $\mathbb Z^{n-1}$. Exactness gives $H_2(T)=0$ and a <short exact sequence>
$$
0\longrightarrow\mathbb Z^{n-1}\longrightarrow H_1(T)\longrightarrow H_1(S)\cong\mathbb Z^{2g}\longrightarrow0.
$$
This sequence splits because its right-hand group is a <free abelian group>. Since $T$ is connected, we obtain
$$
\boxed{H_i(T;\mathbb Z)\cong
\begin{cases}
\mathbb Z,&i=0,\\
\mathbb Z^{2g+n-1},&i=1,\\
0,&\text{otherwise}.
\end{cases}}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 21F / b / Solution

For a pair $A\subseteq X$, the <short exact sequence of chain complexes> $0\to C_*(A)\to C_*(X)\to C_*(X,A)\to0$ induces
$$
\cdots\to H_n(A)\to H_n(X)\to H_n(X,A)\xrightarrow{\partial}H_{n-1}(A)\to\cdots.
$$

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 Incoming links: Long exact sequence in relative homology