[proofplan]
We construct an explicit isomorphism $H_0(X) \cong \mathbb{Z}$ when $X$ is path-connected and nonempty. The augmentation map $\varepsilon: C_0(X) \to \mathbb{Z}$ sends each $0$-chain to the sum of its coefficients. We show $\varepsilon$ vanishes on boundaries (so it descends to homology), is surjective (since $X$ is nonempty), and is injective on $H_0(X)$ (since path-connectedness implies every difference of two points is a boundary).
[/proofplan]
[step:Define the augmentation homomorphism and verify it descends to $H_0$]
Define the augmentation homomorphism
\begin{align*}
\varepsilon: C_0(X) &\to \mathbb{Z} \\
\sum_{\sigma} n_\sigma \, \sigma &\mapsto \sum_{\sigma} n_\sigma,
\end{align*}
where the sum runs over singular $0$-simplices $\sigma: \Delta^0 \to X$ (i.e., points of $X$) and only finitely many $n_\sigma$ are nonzero. This is a well-defined homomorphism.
We verify that $\varepsilon$ vanishes on the image of $d_1: C_1(X) \to C_0(X)$. Let $\tau: \Delta^1 \to X$ be a singular $1$-simplex. Then
\begin{align*}
d_1(\tau) = \tau \circ \delta_0 - \tau \circ \delta_1,
\end{align*}
where $\tau \circ \delta_0$ and $\tau \circ \delta_1$ are singular $0$-simplices. Applying $\varepsilon$:
\begin{align*}
\varepsilon(d_1(\tau)) = 1 - 1 = 0.
\end{align*}
Since $\varepsilon$ vanishes on every generator of $\operatorname{im}(d_1) = B_0(X)$, it vanishes on all of $B_0(X)$. Hence $\varepsilon$ descends to a well-defined homomorphism $\bar{\varepsilon}: H_0(X) = C_0(X)/B_0(X) \to \mathbb{Z}$ (noting that $H_0(X) = C_0(X)/\operatorname{im}(d_1)$ since $C_{-1}(X) = 0$ means every $0$-chain is a cycle).
[guided]
Why is every $0$-chain a $0$-cycle? Because $d_0: C_0(X) \to C_{-1}(X) = 0$ is the zero map. So $Z_0(X) = \ker(d_0) = C_0(X)$, and $H_0(X) = Z_0(X)/B_0(X) = C_0(X)/\operatorname{im}(d_1)$.
The augmentation map $\varepsilon$ simply "forgets the geometry" — it remembers only the total coefficient, ignoring which points the $0$-simplices map to. The verification $\varepsilon(d_1(\tau)) = 0$ says that the boundary of any $1$-simplex contributes coefficient $+1$ at one endpoint and $-1$ at the other, which sums to zero.
[/guided]
[/step]
[step:Verify surjectivity of $\bar{\varepsilon}$]
Since $X$ is nonempty, there exists a point $x_0 \in X$. The constant map $\sigma_0: \Delta^0 \to X$ with $\sigma_0(\Delta^0) = \{x_0\}$ is a singular $0$-simplex, and $\varepsilon(\sigma_0) = 1$. Hence $\bar{\varepsilon}([\sigma_0]) = 1$, and since $\mathbb{Z}$ is generated by $1$, the map $\bar{\varepsilon}$ is surjective.
[/step]
[step:Verify injectivity of $\bar{\varepsilon}$ using path-connectedness]
Let $c = \sum_{k=1}^{m} n_k \sigma_k \in C_0(X)$ be a $0$-chain with $\varepsilon(c) = \sum_k n_k = 0$. We must show $c \in B_0(X)$, i.e., $c$ is a boundary.
Fix a basepoint $x_0 \in X$ and write $\sigma_0$ for the constant $0$-simplex at $x_0$. For each $k$, let $p_k = \sigma_k(\Delta^0) \in X$ be the point that $\sigma_k$ maps to. Since $X$ is path-connected, there exists a continuous path $\gamma_k: [0,1] \to X$ with $\gamma_k(0) = x_0$ and $\gamma_k(1) = p_k$.
The path $\gamma_k$ defines a singular $1$-simplex $\tau_k: \Delta^1 \to X$ (identifying $\Delta^1 \cong [0,1]$), and its boundary is
\begin{align*}
d_1(\tau_k) = \sigma_k - \sigma_0.
\end{align*}
Therefore $\sigma_k = \sigma_0 + d_1(\tau_k)$ in $C_0(X)$, so $\sigma_k \equiv \sigma_0 \pmod{B_0(X)}$. Substituting:
\begin{align*}
c = \sum_{k=1}^{m} n_k \sigma_k \equiv \sum_{k=1}^{m} n_k \sigma_0 = \left(\sum_{k=1}^{m} n_k\right) \sigma_0 = 0 \cdot \sigma_0 = 0 \pmod{B_0(X)},
\end{align*}
where the last equality uses $\varepsilon(c) = \sum_k n_k = 0$. Hence $[c] = 0$ in $H_0(X)$, and $\bar{\varepsilon}$ is injective.
[guided]
The key idea is that path-connectedness makes all points "homologous." If $p$ and $q$ are two points in $X$, any path from $p$ to $q$ defines a singular $1$-simplex $\tau$ with $d_1(\tau) = q - p$ (viewing $p$ and $q$ as $0$-simplices). This means $[p] = [q]$ in $H_0(X)$.
So in $H_0(X)$, every $0$-simplex represents the same class $[\sigma_0]$. A general $0$-chain $c = \sum n_k \sigma_k$ therefore satisfies $[c] = (\sum n_k)[\sigma_0] = \varepsilon(c) \cdot [\sigma_0]$. The map $\bar{\varepsilon}$ sends $[c]$ to $\varepsilon(c)$, and this formula shows $\bar{\varepsilon}$ is injective: $\varepsilon(c) = 0$ implies $[c] = 0$.
What if $X$ were not path-connected? Then distinct path components would contribute independent generators of $H_0(X)$, and we would get $H_0(X) \cong \mathbb{Z}^{\pi_0(X)}$, where $\pi_0(X)$ is the set of path components.
[/guided]
[/step]
[step:Conclude the isomorphism $H_0(X) \cong \mathbb{Z}$]
Since $\bar{\varepsilon}: H_0(X) \to \mathbb{Z}$ is a surjective and injective homomorphism, it is an isomorphism. Hence $H_0(X) \cong \mathbb{Z}$.
[/step]
