[proofplan] The proof is a direct application of the universal coefficient theorem combined with two elementary tensor-product computations over $\mathbb{Z}$: first, $\mathbb{Z}^k \otimes_\mathbb{Z} \mathbb{Q} \cong \mathbb{Q}^k$, and second, $F \otimes_\mathbb{Z} \mathbb{Q} = 0$ for every finite abelian group $F$. Since $\mathbb{Q}$ is a flat $\mathbb{Z}$-module (equivalently, since $\operatorname{Tor}_1^\mathbb{Z}(A, \mathbb{Q}) = 0$ for every abelian group $A$), rational homology coincides with the base change $H_n(K) \otimes_\mathbb{Z} \mathbb{Q}$, and the computation decomposes across the direct-sum splitting $H_n(K) \cong \mathbb{Z}^k \oplus F$ to yield $\mathbb{Q}^k$. [/proofplan] [step:Identify $H_n(K; \mathbb{Q})$ with the rationalisation $H_n(K) \otimes_\mathbb{Z} \mathbb{Q}$] Let $C_\bullet(K)$ denote the integral simplicial chain complex of $K$. By definition, the rational chain complex is $C_\bullet(K; \mathbb{Q}) = C_\bullet(K) \otimes_\mathbb{Z} \mathbb{Q}$, where the boundary maps are extended $\mathbb{Q}$-linearly. The universal coefficient theorem for homology, applied to the coefficient ring $\mathbb{Q}$, produces a natural short exact sequence \begin{align*} 0 \to H_n(K) \otimes_\mathbb{Z} \mathbb{Q} \to H_n(K; \mathbb{Q}) \to \operatorname{Tor}_1^\mathbb{Z}(H_{n-1}(K), \mathbb{Q}) \to 0. \end{align*} Since $\mathbb{Q}$ is a flat $\mathbb{Z}$-module (every short exact sequence of abelian groups remains exact after $\otimes_\mathbb{Z} \mathbb{Q}$, because localisation is exact), $\operatorname{Tor}_1^\mathbb{Z}(A, \mathbb{Q}) = 0$ for every abelian group $A$. In particular the right-hand term vanishes, and the first map is an isomorphism: \begin{align*} H_n(K; \mathbb{Q}) \cong H_n(K) \otimes_\mathbb{Z} \mathbb{Q}. \end{align*} [guided] There are two ways to define $H_n(K; \mathbb{Q})$: (i) as the homology of the rational chain complex $C_\bullet(K) \otimes_\mathbb{Z} \mathbb{Q}$, which is how the definition was stated in the course; and (ii) as the base change $H_n(K) \otimes_\mathbb{Z} \mathbb{Q}$. These are not a priori the same — taking homology and tensoring do not commute in general, and the discrepancy is measured by $\operatorname{Tor}$. Concretely, the universal coefficient theorem gives the natural short exact sequence \begin{align*} 0 \to H_n(K) \otimes_\mathbb{Z} \mathbb{Q} \to H_n(K; \mathbb{Q}) \to \operatorname{Tor}_1^\mathbb{Z}(H_{n-1}(K), \mathbb{Q}) \to 0. \end{align*} Why does $\operatorname{Tor}$ enter? Tensoring a chain complex by $\mathbb{Q}$ preserves kernels of the boundary maps (cycles go to cycles because $d \otimes \operatorname{id}_\mathbb{Q}$ still vanishes on them), but may enlarge or shrink the image (boundaries); the discrepancy is the $\operatorname{Tor}$ term. The flatness of $\mathbb{Q}$ over $\mathbb{Z}$ kills this discrepancy. One way to see $\operatorname{Tor}_1^\mathbb{Z}(A, \mathbb{Q}) = 0$ for all abelian groups $A$: take any free resolution $0 \to F_1 \to F_0 \to A \to 0$ of $A$ (which exists because $\mathbb{Z}$ is a PID, so every abelian group has a length-1 free resolution). Tensoring with $\mathbb{Q}$ gives $0 \to F_1 \otimes \mathbb{Q} \to F_0 \otimes \mathbb{Q} \to A \otimes \mathbb{Q} \to 0$; the sequence remains short exact because $\mathbb{Q} = \operatorname{colim}_{n \geq 1} \frac{1}{n}\mathbb{Z}$ is a filtered colimit of free $\mathbb{Z}$-modules (each copy of $\frac{1}{n}\mathbb{Z} \subset \mathbb{Q}$ is free of rank $1$), hence flat. So $\operatorname{Tor}_1^\mathbb{Z}(A, \mathbb{Q}) = \ker(F_1 \otimes \mathbb{Q} \to F_0 \otimes \mathbb{Q}) = 0$. With the $\operatorname{Tor}$ term gone, the universal coefficient sequence collapses to an isomorphism $H_n(K; \mathbb{Q}) \cong H_n(K) \otimes_\mathbb{Z} \mathbb{Q}$. From this point on, the computation is purely algebraic. [/guided] [/step] [step:Compute $\mathbb{Z}^k \otimes_\mathbb{Z} \mathbb{Q} \cong \mathbb{Q}^k$ using additivity of tensor product] The tensor product distributes over direct sums: for any $\mathbb{Z}$-modules $A_1, \ldots, A_r$ and $B$, \begin{align*} (A_1 \oplus \cdots \oplus A_r) \otimes_\mathbb{Z} B \cong (A_1 \otimes_\mathbb{Z} B) \oplus \cdots \oplus (A_r \otimes_\mathbb{Z} B). \end{align*} Applying this to $\mathbb{Z}^k = \mathbb{Z} \oplus \cdots \oplus \mathbb{Z}$ ($k$ summands) with $B = \mathbb{Q}$ and using $\mathbb{Z} \otimes_\mathbb{Z} \mathbb{Q} \cong \mathbb{Q}$ (the unit isomorphism of the tensor product), we obtain \begin{align*} \mathbb{Z}^k \otimes_\mathbb{Z} \mathbb{Q} \cong \mathbb{Q}^k. \end{align*} [/step] [step:Show $F \otimes_\mathbb{Z} \mathbb{Q} = 0$ for every finite abelian group $F$] Let $F$ be a finite abelian group. By the structure theorem for finitely generated abelian groups, $F$ decomposes as a finite direct sum of cyclic groups of finite order: \begin{align*} F \cong \bigoplus_{j=1}^{s} \mathbb{Z}/n_j\mathbb{Z}, \qquad n_j \geq 2. \end{align*} By additivity of the tensor product over direct sums, it suffices to show $(\mathbb{Z}/n\mathbb{Z}) \otimes_\mathbb{Z} \mathbb{Q} = 0$ for every integer $n \geq 1$. Fix $n \geq 1$ and let $\bar{1} \in \mathbb{Z}/n\mathbb{Z}$ denote the generator; every element of $\mathbb{Z}/n\mathbb{Z}$ is a $\mathbb{Z}$-multiple of $\bar{1}$. Then for any $q \in \mathbb{Q}$, using $\mathbb{Z}$-bilinearity of the tensor and the fact that $n \cdot \bar{1} = 0$ in $\mathbb{Z}/n\mathbb{Z}$, \begin{align*} \bar{1} \otimes q = \bar{1} \otimes \left(n \cdot \frac{q}{n}\right) = (n \cdot \bar{1}) \otimes \frac{q}{n} = 0 \otimes \frac{q}{n} = 0. \end{align*} (Here we used that $\mathbb{Q}$ is divisible: every rational $q$ has the form $n \cdot (q/n)$, so dividing by $n$ stays within $\mathbb{Q}$.) Hence every simple tensor $\bar{1} \otimes q$ vanishes. Since $\bar{1}$ generates $\mathbb{Z}/n\mathbb{Z}$ and simple tensors generate $(\mathbb{Z}/n\mathbb{Z}) \otimes_\mathbb{Z} \mathbb{Q}$, the whole tensor product is zero. Summing over $j$, $F \otimes_\mathbb{Z} \mathbb{Q} = 0$. [guided] The core observation is that $\mathbb{Z}/n\mathbb{Z}$ is $n$-torsion (every element is killed by multiplication by $n$), while $\mathbb{Q}$ is $n$-divisible (every element is divisible by $n$). In the tensor product, we can slide scalars across the $\otimes$ symbol: for any $a \in \mathbb{Z}/n\mathbb{Z}$, $q \in \mathbb{Q}$, and $m \in \mathbb{Z}$, \begin{align*} a \otimes q = a \otimes \left(m \cdot \frac{q}{m}\right) = (ma) \otimes \frac{q}{m}, \qquad \text{provided } q/m \in \mathbb{Q}. \end{align*} Taking $m = n$ and $a = \bar{1}$: $\bar{1} \otimes q = (n \cdot \bar{1}) \otimes (q/n) = 0 \otimes (q/n) = 0$. Why does this not work for $F \otimes_\mathbb{Z} \mathbb{Z}$? Because $\mathbb{Z}$ is not divisible — we cannot form $q/n$ inside $\mathbb{Z}$. This is the precise mechanism by which passing from integer to rational coefficients kills torsion: the torsion-divisibility mismatch annihilates the tensor product. The extension from $\mathbb{Z}/n\mathbb{Z}$ to a general finite abelian group is immediate: the structure theorem reduces $F$ to a direct sum of finite cyclic groups, and additivity of the tensor product reduces the sum to the cyclic case. [/guided] [/step] [step:Combine the computations to conclude $H_n(K; \mathbb{Q}) \cong \mathbb{Q}^k$] By hypothesis $H_n(K) \cong \mathbb{Z}^k \oplus F$ with $F$ a finite abelian group. Combining the previous steps: \begin{align*} H_n(K; \mathbb{Q}) &\cong H_n(K) \otimes_\mathbb{Z} \mathbb{Q} && \text{(Step 1, universal coefficient + flatness)} \\ &\cong (\mathbb{Z}^k \oplus F) \otimes_\mathbb{Z} \mathbb{Q} && \text{(hypothesis)} \\ &\cong (\mathbb{Z}^k \otimes_\mathbb{Z} \mathbb{Q}) \oplus (F \otimes_\mathbb{Z} \mathbb{Q}) && \text{(additivity of } \otimes \text{)} \\ &\cong \mathbb{Q}^k \oplus 0 && \text{(Steps 2 and 3)} \\ &\cong \mathbb{Q}^k. \end{align*} This completes the proof. [/step]