Rotman’s proof of $H_1(X,x_0) \cong H_1(X)$

Question

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

Answer 1

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.