[proofplan] We use the long exact sequence of the pair $(X, \{x_0\})$ to relate the ordinary homology $H_n(X)$ to the relative homology $H_n(X, \{x_0\})$, which we identify with the reduced homology $\tilde{H}_n(X)$. For $n \geq 1$, the vanishing of $H_n(\{x_0\})$ and $H_{n-1}(\{x_0\})$ gives an isomorphism $H_n(X) \cong \tilde{H}_n(X)$ directly. For $n = 0$, the sequence yields a short exact sequence that splits because the inclusion $\{x_0\} \hookrightarrow X$ provides a section. [/proofplan] [step:Write out the long exact sequence of the pair $(X, \{x_0\})$] The pair $(X, \{x_0\})$ gives rise to a long exact sequence in homology: \begin{align*} \cdots \to H_n(\{x_0\}) \xrightarrow{i_*} H_n(X) \xrightarrow{q_*} H_n(X, \{x_0\}) \xrightarrow{\partial} H_{n-1}(\{x_0\}) \to \cdots \end{align*} where $i: \{x_0\} \hookrightarrow X$ is the inclusion and $q: (X, \varnothing) \to (X, \{x_0\})$ is the map of pairs. We identify $\tilde{H}_n(X) := H_n(X, \{x_0\})$ (the relative homology of the pair is one standard definition of reduced homology; it agrees with the kernel-of-augmentation definition). The homology of a one-point space is \begin{align*} H_n(\{x_0\}) = \begin{cases} \mathbb{Z} & n = 0, \\ 0 & n \geq 1. \end{cases} \end{align*} [guided] The long exact sequence of a pair is the fundamental tool for relating the homology of a space, a subspace, and their relative homology. Here the subspace is the single point $\{x_0\}$, which has particularly simple homology: $H_0(\{x_0\}) \cong \mathbb{Z}$ (generated by the unique $0$-simplex at $x_0$) and $H_n(\{x_0\}) = 0$ for $n \geq 1$ (a point has no higher-dimensional cycles). The reduced homology $\tilde{H}_n(X)$ is defined as $H_n(X, \{x_0\})$. Equivalently, $\tilde{H}_n(X)$ can be defined as the homology of the augmented chain complex $\cdots \to C_1(X) \xrightarrow{d_1} C_0(X) \xrightarrow{\varepsilon} \mathbb{Z} \to 0$, where $\varepsilon(\sum n_\sigma \sigma) = \sum n_\sigma$. The two definitions agree because the short exact sequence $0 \to \tilde{C}_\bullet(\{x_0\}) \to \tilde{C}_\bullet(X) \to C_\bullet(X, \{x_0\}) \to 0$ of augmented chain complexes gives the same long exact sequence, and the augmented complex of a point is exact. [/guided] [/step] [step:Establish the isomorphism $H_n(X) \cong \tilde{H}_n(X)$ for $n \geq 1$] For $n \geq 2$, the relevant portion of the long exact sequence reads \begin{align*} H_n(\{x_0\}) \xrightarrow{i_*} H_n(X) \xrightarrow{q_*} \tilde{H}_n(X) \xrightarrow{\partial} H_{n-1}(\{x_0\}). \end{align*} Since $H_n(\{x_0\}) = 0$ and $H_{n-1}(\{x_0\}) = 0$ for $n \geq 2$, exactness gives: $q_*$ is injective (its kernel equals $\operatorname{im}(i_*) = 0$) and surjective (its cokernel is $\ker(\partial) / \operatorname{im}(q_*)$, but $\partial$ is zero since $H_{n-1}(\{x_0\}) = 0$, and surjectivity follows from exactness at $\tilde{H}_n(X)$: $\ker(\partial) = \operatorname{im}(q_*)$, with $\ker(\partial) = \tilde{H}_n(X)$). Therefore $q_*: H_n(X) \xrightarrow{\sim} \tilde{H}_n(X)$ is an isomorphism for $n \geq 2$. For $n = 1$, the sequence reads \begin{align*} H_1(\{x_0\}) \xrightarrow{i_*} H_1(X) \xrightarrow{q_*} \tilde{H}_1(X) \xrightarrow{\partial} H_0(\{x_0\}) \xrightarrow{i_*} H_0(X). \end{align*} We have $H_1(\{x_0\}) = 0$, so $q_*$ is injective. For surjectivity, we examine the map $\partial: \tilde{H}_1(X) \to H_0(\{x_0\}) \cong \mathbb{Z}$. The subsequent map $i_*: H_0(\{x_0\}) \to H_0(X)$ sends the generator $[x_0] \in H_0(\{x_0\})$ to $[x_0] \in H_0(X)$, which is a nonzero class (since $X$ is nonempty, $H_0(X) \neq 0$ and $i_*$ is injective on $H_0$ when the basepoint lies in a path component). In general, $i_*$ need not be injective, but exactness at $H_0(\{x_0\})$ gives $\ker(i_*) = \operatorname{im}(\partial)$. However, $i_*: H_0(\{x_0\}) \to H_0(X)$ is always injective: the class $[x_0]$ is nonzero in $H_0(X)$ because the augmentation $\varepsilon: H_0(X) \to \mathbb{Z}$ satisfies $\varepsilon([x_0]) = 1 \neq 0$. Therefore $\operatorname{im}(\partial) = \ker(i_*) = 0$, so $\partial = 0$ and $q_*$ is surjective. Hence $q_*: H_1(X) \xrightarrow{\sim} \tilde{H}_1(X)$. [guided] For $n \geq 2$, both the term before and the term after $q_*$ vanish, so $q_*$ is sandwiched between zeros in the exact sequence, forcing it to be an isomorphism. This is a standard "two zeros" argument. The case $n = 1$ requires more care because $H_0(\{x_0\}) \cong \mathbb{Z} \neq 0$ appears after $\partial$. We need to show $\partial: \tilde{H}_1(X) \to H_0(\{x_0\})$ is zero. By exactness, $\operatorname{im}(\partial) = \ker(i_*: H_0(\{x_0\}) \to H_0(X))$. Is $i_*$ injective on $H_0$? Yes. The generator of $H_0(\{x_0\}) \cong \mathbb{Z}$ is the class $[x_0]$. Its image $i_*([x_0]) = [x_0] \in H_0(X)$ is nonzero because the augmentation map $\varepsilon: H_0(X) \to \mathbb{Z}$, defined by $\varepsilon(\sum n_\sigma \sigma) = \sum n_\sigma$, satisfies $\varepsilon([x_0]) = 1$. Since $\varepsilon \circ i_*$ is nonzero, $i_*$ must be injective (it is a homomorphism $\mathbb{Z} \to H_0(X)$ sending the generator to a nonzero element). Therefore $\ker(i_*) = 0$, so $\operatorname{im}(\partial) = 0$, so $\partial = 0$, and $q_*: H_1(X) \to \tilde{H}_1(X)$ is both injective (since $H_1(\{x_0\}) = 0$) and surjective (since $\ker(\partial) = \tilde{H}_1(X) = \operatorname{im}(q_*)$). [/guided] [/step] [step:Derive the split short exact sequence for $n = 0$] For $n = 0$, the relevant portion of the long exact sequence is \begin{align*} H_0(\{x_0\}) \xrightarrow{i_*} H_0(X) \xrightarrow{q_*} \tilde{H}_0(X) \to 0, \end{align*} where the zero on the right comes from $H_{-1}(\{x_0\}) = 0$ (there is no homology in negative degrees). Thus $q_*$ is surjective. By exactness, $\ker(q_*) = \operatorname{im}(i_*)$. Since $i_*$ is injective (as shown above), we obtain the short exact sequence \begin{align*} 0 \to \mathbb{Z} \xrightarrow{i_*} H_0(X) \xrightarrow{q_*} \tilde{H}_0(X) \to 0. \end{align*} This sequence splits. A splitting is provided by the section $s: \mathbb{Z} \to H_0(X)$ that sends $1 \mapsto [x_0]$, which equals $i_*$. Equivalently, the augmentation $\varepsilon: H_0(X) \to \mathbb{Z}$ defined by $\varepsilon([x]) = 1$ for any point $x$ satisfies $\varepsilon \circ i_* = \operatorname{id}_\mathbb{Z}$, so $i_*$ is a section of $\varepsilon$. By the splitting lemma for abelian groups, \begin{align*} H_0(X) \cong \tilde{H}_0(X) \oplus \mathbb{Z}. \end{align*} [guided] In degree zero, we cannot get an isomorphism between $H_0$ and $\tilde{H}_0$ because the inclusion of the basepoint contributes a copy of $\mathbb{Z}$ to $H_0(X)$. The reduced homology $\tilde{H}_0(X)$ "quotients out" this basepoint contribution. The short exact sequence $0 \to \mathbb{Z} \xrightarrow{i_*} H_0(X) \xrightarrow{q_*} \tilde{H}_0(X) \to 0$ tells us that $H_0(X)$ is an extension of $\tilde{H}_0(X)$ by $\mathbb{Z}$. Does this extension split? Yes. The augmentation map $\varepsilon: H_0(X) \to \mathbb{Z}$, which sends any $0$-cycle $\sum n_\sigma \sigma$ to $\sum n_\sigma$, provides a left inverse (retraction) to $i_*$: $\varepsilon(i_*([x_0])) = \varepsilon([x_0]) = 1$. So $\varepsilon \circ i_* = \operatorname{id}_\mathbb{Z}$. By the splitting lemma, whenever a short exact sequence $0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0$ of abelian groups admits a retraction $r: B \to A$ with $r \circ f = \operatorname{id}_A$, we have $B \cong A \oplus C$ via $b \mapsto (r(b), g(b))$. Here $A = \mathbb{Z}$, $B = H_0(X)$, $C = \tilde{H}_0(X)$, and $r = \varepsilon$. Concretely, for a space with $m$ path components, $H_0(X) \cong \mathbb{Z}^m$ and $\tilde{H}_0(X) \cong \mathbb{Z}^{m-1}$, confirming $H_0(X) \cong \tilde{H}_0(X) \oplus \mathbb{Z}$. [/guided] [/step]