TITLE: Rotman's proof of $H_1(X,x_0) \cong H_1(X)$
QUESTION [8 upvotes]: 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.
(Original scanned images here and here.)

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 
$$
      H_0(X)
\cong H_0(\{ x_0\}) \oplus H_0(X,x_0)
\cong \mathbb Z\oplus H_0(X,x_0).
$$

REPLY [9 votes]: 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.