[proofplan] We prove all three parts using the structure theory of CW complexes. For part (1), we use the good pair property of $(X^n, X^{n-1})$ to reduce to the reduced homology of the wedge $X^n/X^{n-1} \cong \bigvee_\alpha S^n_\alpha$, then compute the homology of a wedge of spheres. For part (2), we induct downward on the skeleton dimension, using part (1) and the long exact sequence of the pair $(X^m, X^{m-1})$. For part (3), we induct upward, showing that attaching cells of dimension $> i + 1$ does not affect $H_i$. [/proofplan] [step:Prove part (1): compute $H_i(X^n, X^{n-1})$] The pair $(X^n, X^{n-1})$ is a good pair: $X^{n-1}$ is a subcomplex of $X^n$, and each $n$-cell $e^n_\alpha$ has a characteristic map $\Phi_\alpha: D^n \to X^n$ with $\Phi_\alpha(\partial D^n) \subset X^{n-1}$, from which one can construct an open neighbourhood of $X^{n-1}$ in $X^n$ that deformation-retracts onto $X^{n-1}$ (by retracting a collar of each $n$-cell toward its boundary). By the [Quotient Isomorphism for Good Pairs](/theorems/2258), \begin{align*} H_i(X^n, X^{n-1}) \cong \tilde{H}_i(X^n / X^{n-1}). \end{align*} The quotient $X^n / X^{n-1}$ is obtained by collapsing $X^{n-1}$ to a point. Each $n$-cell $e^n_\alpha$ has its boundary in $X^{n-1}$ collapsed to the basepoint, so the cell becomes a copy of $D^n / \partial D^n \cong S^n$. The resulting space is a wedge of $n$-spheres: \begin{align*} X^n / X^{n-1} \cong \bigvee_{\alpha \in I_n} S^n_\alpha, \end{align*} where $I_n$ is the index set of $n$-cells. [claim:The reduced homology of a wedge of spheres is the direct sum of the reduced homologies of the summands] For a finite or infinite wedge $\bigvee_{\alpha \in I} S^n_\alpha$, we have $\tilde{H}_i\left(\bigvee_\alpha S^n_\alpha\right) \cong \bigoplus_\alpha \tilde{H}_i(S^n_\alpha)$. [/claim] [proof] For a finite wedge, we proceed by induction on the number of summands. The base case (one sphere) is immediate. For the inductive step, let $Y = \bigvee_{\alpha=1}^{k-1} S^n_\alpha$ and consider the good pair $(Y \vee S^n_k, \{*\})$. The inclusion of each summand $S^n_\alpha \hookrightarrow \bigvee S^n_\alpha$ induces a map on reduced homology, and the Mayer-Vietoris sequence for the decomposition (choosing open sets that deformation-retract onto the summands with contractible intersection) gives \begin{align*} \tilde{H}_i\left(\bigvee_{\alpha=1}^{k} S^n_\alpha\right) \cong \tilde{H}_i(Y) \oplus \tilde{H}_i(S^n_k) \cong \bigoplus_{\alpha=1}^{k} \tilde{H}_i(S^n_\alpha). \end{align*} For an infinite wedge $\bigvee_{\alpha \in I} S^n_\alpha$ with the weak topology (a CW complex with one $0$-cell and one $n$-cell for each $\alpha$), singular homology commutes with direct limits of CW subcomplexes: $\tilde{H}_i(\bigvee_\alpha S^n_\alpha) = \varinjlim_{F \subset I, \, F \text{ finite}} \tilde{H}_i(\bigvee_{\alpha \in F} S^n_\alpha) = \bigoplus_{\alpha \in I} \tilde{H}_i(S^n_\alpha)$. [/proof] Since $\tilde{H}_i(S^n) \cong \mathbb{Z}$ for $i = n$ and $\tilde{H}_i(S^n) = 0$ for $i \neq n$ (by the [Homology of Spheres](/theorems/1945)), we obtain \begin{align*} H_i(X^n, X^{n-1}) \cong \tilde{H}_i\left(\bigvee_{\alpha \in I_n} S^n_\alpha\right) \cong \begin{cases} \bigoplus_{\alpha \in I_n} \mathbb{Z} & i = n, \\ 0 & i \neq n. \end{cases} \end{align*} [guided] The key idea is that collapsing $X^{n-1}$ to a point "frees" each $n$-cell: its boundary (which was attached to $X^{n-1}$) gets collapsed to the basepoint, turning the cell $D^n / \partial D^n$ into a sphere $S^n$. Since the cells only interact through their boundaries (which are all now at the basepoint), the resulting space is a wedge of spheres. Why is $(X^n, X^{n-1})$ a good pair? A CW pair $(Y, A)$ where $A$ is a subcomplex is always a good pair. The deformation retract neighbourhood is constructed as follows: for each $n$-cell $e^n_\alpha$ with characteristic map $\Phi_\alpha: D^n \to X^n$, the open collar $\Phi_\alpha(\{x \in D^n : |x| > 1 - \varepsilon\})$ deformation-retracts onto $\Phi_\alpha(\partial D^n) \subset X^{n-1}$ via the radial map $x \mapsto x / |x|$. The union of $X^{n-1}$ with these collars is an open neighbourhood of $X^{n-1}$ in $X^n$ that deformation-retracts onto $X^{n-1}$. The computation of $\tilde{H}_i(\bigvee S^n_\alpha)$ uses the fact that the wedge basepoint is a good subspace (each summand is a CW complex with the basepoint as a $0$-cell), so Mayer-Vietoris or the direct sum decomposition applies. The result is the direct sum $\bigoplus_\alpha \tilde{H}_i(S^n_\alpha)$, which is $\bigoplus_\alpha \mathbb{Z}$ in degree $n$ and zero otherwise. [/guided] [/step] [step:Prove part (2): $H_i(X^n) = 0$ for $i > n$] We prove this by downward induction on $n$. The base case is: if $X$ has no cells above dimension $N$ (i.e., $X = X^N$), then the statement holds vacuously for $n = N$ since there is nothing to prove (we need $H_i(X^N) = 0$ for $i > N$, which follows from the argument below). For the inductive step, consider the long exact sequence of the pair $(X^n, X^{n-1})$: \begin{align*} H_i(X^{n-1}) \xrightarrow{j_*} H_i(X^n) \xrightarrow{q_*} H_i(X^n, X^{n-1}) \xrightarrow{\partial} H_{i-1}(X^{n-1}). \end{align*} By part (1), $H_i(X^n, X^{n-1}) = 0$ for $i \neq n$. When $i > n$, we have $i \neq n$ and also $i - 1 \neq n$ (since $i > n$ implies $i - 1 \geq n$, and if $i - 1 = n$ then $i = n + 1$, but $H_{n+1}(X^n, X^{n-1}) = 0$ since $n + 1 \neq n$). Therefore the sequence gives \begin{align*} H_i(X^{n-1}) \xrightarrow{j_*} H_i(X^n) \to 0, \end{align*} so $j_*$ is surjective for $i > n$. Applying this repeatedly: for $i > n$, \begin{align*} H_i(X^n) \twoheadleftarrow H_i(X^{n-1}) \twoheadleftarrow \cdots \twoheadleftarrow H_i(X^0). \end{align*} Since $X^0$ is a discrete set of points, $H_i(X^0) = 0$ for $i \geq 1$. In particular, $H_i(X^0) = 0$ for $i > n \geq 0$, and the surjections give $H_i(X^n) = 0$. [guided] The idea is that the $n$-skeleton $X^n$ is built from $X^0$ by attaching cells of dimension $\leq n$. The long exact sequence of a pair $(X^m, X^{m-1})$ tells us how $H_i$ changes when we attach $m$-cells. If $i > m$ (the cells are too low-dimensional to create $i$-cycles), the relative term $H_i(X^m, X^{m-1})$ vanishes, and the inclusion map $H_i(X^{m-1}) \to H_i(X^m)$ is surjective. Starting from $H_i(X^0) = 0$ (a discrete space has no higher homology) and surjecting upward through $X^1, X^2, \ldots, X^n$, we get $H_i(X^n) = 0$. More precisely, for $i > n$: - $H_i(X^0) = 0$ (discrete set). - $H_i(X^0) \twoheadrightarrow H_i(X^1)$: here $i > n \geq 1$, and $H_i(X^1, X^0) = 0$ since $i \neq 1$. So actually the map goes the wrong way in the exact sequence. Let us be more careful: the exact sequence gives $H_i(X^{m-1}) \to H_i(X^m) \to H_i(X^m, X^{m-1})$. Since $H_i(X^m, X^{m-1}) = 0$ for $i \neq m$, the map $j_*: H_i(X^{m-1}) \to H_i(X^m)$ is surjective when $i \neq m$ (and also injective when $i \neq m$ and $i - 1 \neq m$, but surjectivity suffices). For $i > n$ and $1 \leq m \leq n$, we have $i > n \geq m$, so $i \neq m$, and the surjectivity gives a chain of surjections $H_i(X^0) \twoheadrightarrow H_i(X^1) \twoheadrightarrow \cdots \twoheadrightarrow H_i(X^n)$. Since $H_i(X^0) = 0$, every group in the chain is zero. [/guided] [/step] [step:Prove part (3): $H_i(X^n) \xrightarrow{\sim} H_i(X)$ for $i < n$] We show that for $m > n$ and $i < n$, the inclusion-induced map $H_i(X^{m-1}) \to H_i(X^m)$ is an isomorphism. From the long exact sequence of $(X^m, X^{m-1})$: \begin{align*} H_{i+1}(X^m, X^{m-1}) \to H_i(X^{m-1}) \xrightarrow{j_*} H_i(X^m) \to H_i(X^m, X^{m-1}). \end{align*} For $i < n < m$, we have $i \neq m$ and $i + 1 \leq n < m$, so $i + 1 \neq m$ as well. By part (1), both $H_{i+1}(X^m, X^{m-1}) = 0$ and $H_i(X^m, X^{m-1}) = 0$. The exact sequence then gives $j_*: H_i(X^{m-1}) \xrightarrow{\sim} H_i(X^m)$. Applying this for $m = n+1, n+2, \ldots$: \begin{align*} H_i(X^n) \xrightarrow{\sim} H_i(X^{n+1}) \xrightarrow{\sim} H_i(X^{n+2}) \xrightarrow{\sim} \cdots \end{align*} To pass from finite skeleta to $X$ itself: every singular simplex $\sigma: \Delta^i \to X$ has compact image, which is contained in a finite subcomplex of $X$ (by the characteristic property of the weak topology on a CW complex), and hence in some finite skeleton $X^N$. Therefore $H_i(X) = \varinjlim_{m} H_i(X^m)$. Since the maps in the direct system are all isomorphisms for $m \geq n + 1$, the direct limit equals $H_i(X^n)$. [guided] The intuition is that attaching cells of dimension $> i + 1$ cannot create new $i$-cycles (the cells are too high-dimensional) and cannot kill existing $i$-cycles (the boundaries of high-dimensional cells lie in high-dimensional homology). Making this precise: when we attach $m$-cells to form $X^m$ from $X^{m-1}$, the effect on $H_i$ is controlled by the relative group $H_i(X^m, X^{m-1})$. By part (1), this vanishes when $i \neq m$. For $i < n$ and $m > n$, we have $i < n < m$, so both $i$ and $i+1$ are strictly less than $m$. The exact sequence $0 \to H_i(X^{m-1}) \to H_i(X^m) \to 0$ shows the inclusion is an isomorphism. To extend from $H_i(X^n) \cong H_i(X^{n+1}) \cong H_i(X^{n+2}) \cong \cdots$ to $H_i(X^n) \cong H_i(X)$, we use the fact that singular homology of a CW complex equals the direct limit of the homology of its finite skeleta. This follows from the compactness of the standard simplex $\Delta^i$: any singular chain $\sigma: \Delta^i \to X$ has image in some finite skeleton $X^N$ (since the image is compact and each CW complex has the weak topology, compact subsets meet only finitely many cells). Therefore every cycle in $C_i(X)$ is already a cycle in $C_i(X^N)$ for some $N$, and similarly for boundaries. Since all the maps $H_i(X^m) \to H_i(X^{m+1})$ are isomorphisms for $m \geq n$, the direct limit stabilises at $H_i(X^n)$, giving $H_i(X) = \varinjlim_m H_i(X^m) = H_i(X^n)$. [/guided] [/step]