Algebraic Künneth formula

Proof. We may represent the complexes \(A\) and \(B\) up to quasi-isomorphism by bounded below chain complexes \(C\) and \(C'\), respectively, such that the \(R\)-modules \(C_i\) and \(C'_i\) are free for all \(i \in \Z \). Then the derived tensor product \(A \otimes ^{\bL }_R B\) is represented by the algebraic tensor product \(C \otimes _R C'\).

We first treat the special case where all the differentials in \(C\) are zero, so that \(H_i(C) = C_i\) for all \(i\). In this case, the differential on \(C \otimes _R C'\) is given by \(d(c \otimes c') = (-1)^i c \otimes dc'\) for \(c \in C_i\), and the chain complex \(C \otimes _R C'\) decomposes as the direct sum of the complexes \(C_i[i] \otimes _R C'\). Since each \(C_i\) is free, the complex \(C_i \otimes _R C'\) is a direct sum of copies of \(C'\), and hence \[ H_n(C_i[i] \otimes _R C') \;\cong \; C_i \otimes _R H_{n-i}(C') \;=\; H_i(C) \otimes _R H_{n-i}(C'). \] Summing over \(i\) yields an isomorphism \(H_n(C \otimes _R C') \cong \bigoplus _{i+j=n} H_i(C) \otimes _R H_j(C')\), which is the statement of the theorem since the Tor terms vanish (\(H_i(C) = C_i\) being free, hence flat).

For the general case, let \(Z_i \subseteq C_i\) and \(B_i \subseteq C_i\) denote the cycles and boundaries in degree \(i\), respectively. These give subchain complexes \(Z\) and \(B\) of \(C\) with trivial differentials. We have a short exact sequence of chain complexes \[ 0 \to Z \to C \to B[1] \to 0 \] arising from the short exact sequences \(0 \to Z_i \to C_i \to B_{i-1} \to 0\) in each degree. Each of these splits since \(B_{i-1}\) is free, being a submodule of the free module \(C_{i-1}\). Because of this splitting, tensoring with \(C'\) preserves exactness, and we obtain a long exact sequence in homology: \[ \cdots \to H_n(Z \otimes _R C') \to H_n(C \otimes _R C') \to H_{n-1}(B \otimes _R C') \xrightarrow {i_{n-1}} H_{n-1}(Z \otimes _R C') \to \cdots \] where \(i_k\colon H_k(B\otimes _R C')\to H_k(Z\otimes _R C')\) is induced by the inclusion \(B \hookrightarrow Z\).

Since \(Z\) and \(B\) are chain complexes with trivial differentials (and have free terms, being submodules of the free modules \(C_i\)), the special case treated before converts this to: \[ \cdots \xrightarrow {i_n} \bigoplus _{i + j = n} Z_i \otimes _R H_{j}(C') \to H_n(C \otimes _R C') \to \bigoplus _{i+j = n-1} B_i \otimes _R H_{j}(C') \xrightarrow {i_{n-1}} \bigoplus _{i+j = n-1} Z_i \otimes _R H_{j}(C') \to \cdots \] The long exact sequence thus provides short exact sequences \[ 0 \to \coker (i_{n}) \to H_n(C \otimes _R C') \to \ker (i_{n-1}) \to 0. \] To identify \(\coker (i_n)\) and \(\ker (i_{n-1})\), observe that the short exact sequence \(0 \to B_i \to Z_i \to H_i(C) \to 0\) functions as a free resolution of \(H_i(C)\). The long exact sequence on Tor-groups between \(H_i(C)\) and \(H_j(C')\) takes the form \[ 0 \to \Tor _1^R(H_i(C), H_{j}(C')) \to B_i \otimes _R H_{j}(C') \to Z_i \otimes _R H_{j}(C') \to H_i(C) \otimes _R H_{j}(C') \to 0. \] By taking the direct sum over all \(j\), this results in isomorphisms \[ \coker (i_{n}) \;\cong \; \bigoplus _{i+j = n} H_i(C) \otimes _R H_j(C') \qquadtext { and } \ker (i_{n-1}) \;\cong \; \bigoplus _{i + j = n-1} \Tor _1^R(H_i(C), H_j(C')). \] All maps used to construct this short exact sequence are functorial in chain maps. Through the projective-complex model of \(\D ^-(R)\) from Proposition 6.6.16, the sequence is therefore natural in \(A\) and \(B\).

Finally, we show that the short exact sequence splits. Since each \(C_i\) is free, the short exact sequence \(0 \to Z_i \to C_i \to B_{i-1} \to 0\) splits, and the quotient maps \(Z_i \to H_i(C)\) extend to homomorphisms \(C_i \to H_i(C)\). We similarly obtain homomorphisms \(C'_j \to H_j(C')\). Viewing the sequences of homology groups \(H_i(C)\) and \(H_j(C')\) as chain complexes with trivial differentials, we obtain chain maps \(C \to H(C)\) and \(C' \to H(C')\). Their tensor product is a chain map \(C \otimes _R C' \to H(C) \otimes _R H(C')\). Since the differentials on the target are trivial, the induced map on homology \[ H_n(C \otimes _R C') \to H_n(H(C) \otimes _R H(C')) \;=\; \bigoplus _{i+j=n} H_i(C) \otimes _R H_j(C') \] provides the desired splitting. □