Let $X$ be a topological space, and $x_0\in X$. Then $H_n(X)\cong H_n(X,x_0)$ whenever $n\ge 1$. For $n\ge 2,$ it's easy: the exact sequence $$ \cdots \rightarrow H_n(\left \{ x_0 \right \}) \rightarrow H_n(X) \rightarrow H_n(X,x_0) \rightarrow H_{n-1}(\left \{ x_0 \right \}) \rightarrow \cdots $$ and the dimension axiom implies the result whenever $n\ge 2$.
My problem is with Rotman's proof of the case $n=1$, specifically the proof that $\ker k\neq 0$.
\begin{align*} \cdots \to H_1(\{x_0\}) \to H_1(X) \xrightarrow{g} H_1(X,x_0) \to H_0(\{x_0\}) \xrightarrow{h} H_0(X) &\xrightarrow{k} H_0(X,x_0) \\ &\to 0. \end{align*} since $H_1(\{x_0\}) = 0$, the map $g$ is injective; by Exercise 5.2, $g$ is surjective (hence is an isomorphism) if and only if $h$ is injective. The map $h$ has domain $H_0(\{x_0\}) \cong \mathbb{Z}$ and target the free abelian groups $H_0(X)$. If $h \neq 0$, then $h$ must be injective (if $\ker h \neq 0$, then $H_0(X)$ would contain a nontrivial finite subgroup isomorphic to $\mathbb{Z}/{\ker h}$). Now $\operatorname{im} h = \ker k$, so that $\ker k \neq 0$ implies that $\operatorname{im} h \neq 0$, hence $h \neq 0$, as desired. But $k$, being induced by inclusion, is the map $S_0(X)/B_0(X) \to S_0(X)/B_0(X) + S_0(x_0)$ [$S_0(X) = Z_0(X) = Z_0(X,x_0)$] given by $\gamma + B_0(X) \mapsto \gamma + B_0(X) + S_0(x_0)$, and so $\ker k = (B_0(X) + S_0(x_0))/B_0(X)$. The proof of Theorem 4.14 describes $B_0(X)$ as all $\sum m_x x$ with $\sum m_x = 0$; hence $\ker k \neq 0$, and the proof is complete.
Is he making the identification \begin{align*} H(X,x_0) &= \ker \overline{\partial} / {\operatorname{im} \overline{\partial}} \\ &\cong Z_0(X,x_0)) / B_0(X,x_0) \\ &= Z_0(X) / B_0(X,x_0) \\ &= Z_0(X) / ( B_0(X) + S_0(x_0) ) \end{align*} when he claims that $\gamma+B_0(X)\mapsto\gamma +B_0(X)+S_0(x_0)?$
Is there another more intuitive way to treat the case $n=1$?
Edit: or maybe we can just note that if $r \colon X \to \{ x_0 \}$ is the constant map, then $r\circ h = 1_{\{ x_0 \}}$ and therefore $$ 1_{H_0( \{ x_0 \})} = H_0(1_{ \{ x_0 \}}) = H_0(r \circ h) = H_0(r)\circ H_0(h) = r_* \circ h_*, $$ so $h_*$ is injective.
From this we can even get the result that $\tilde H_0 \cong H_0(X,x_0)$, for we know that $$ H_0(X) \cong \mathbb Z \oplus \tilde H_0(X) $$ and, since $h_*$ is injective, $$ 0 \rightarrow H_0(\{ x_0 \}) \xrightarrow{h_*} H_0(X) \xrightarrow{k_*} H_0(X,x_0) \rightarrow 0 $$
is a short exact sequence, and now since $r_*\circ h_*=1$, it splits, so that $$ H_0(X) \cong H_0(\{ x_0\}) \oplus H_0(X,x_0) \cong \mathbb Z\oplus H_0(X,x_0). $$
Rotman's argument is all rather laboured. The point of course is to prove $h$ injective. But $H_0(X)$ is the free Abelian group with generators corresponding to the path components of $X$. To see this it has generators $[x]$ for $x\in X$ and relations $[x]=[y]$ for $x$ and $y$ connected by a path in $X$. So we can pick one $x$ from each path component, throw away all other generators, and get a free generating set.
The map $h$ takes the generator $[x_0]$ of $H_0(\{x_0\})$ to $[x_0]\in H(X)$. As this is one of the free generating set of $H(X)$ discussed above, the map $h$ must be injective.