Künneth theorem

Remarks

For the algebraic part, note that the cycles $Z_\bullet$ and boundaries $B_\bullet$ of $C$ are free, since $R$ is a principal ideal domain. So $0 \to Z_\bullet \to C_\bullet \to B_{\bullet - 1} \to 0$ is a split sequence of free complexes. The long exact sequence after tensoring with $D$ gives the result. The topological part follows from the algebraic one and the Eilenberg–Zilber theorem. That theorem gives a chain homotopy equivalence between the singular chains of $X \times Y$ and the tensor product of the singular chains of $X$ and of $Y$. If $R$ is a field, all Tor terms vanish and $H_n(X \times Y; R) \cong \bigoplus_{i+j=n} H_i(X; R) \otimes_R H_j(Y; R)$. Over $\Z$ the Tor term matters. For the real projective plane $P$, one has $H_0(P) = \Z$, $H_1(P) = \Z/2\Z$ and $H_2(P) = 0$. So the tensor part in degree $3$ is zero, and the formula gives $H_3(P \times P; \Z) \cong \operatorname{Tor}_1^\Z(\Z/2\Z, \Z/2\Z) \cong \Z/2\Z$.

Going deeper

For a student

The theorem is a direct application of the derived functor Tor: the failure of $H(C \otimes D) \cong H(C) \otimes H(D)$ is measured by $\operatorname{Tor}_1$. Taking $D$ to be a module $M$ in degree $0$ gives the homology form of the universal coefficient theorem. There are also cohomology versions, with a cross product into $H^n(X \times Y; R)$, and spectral sequence versions over rings that are not principal ideal domains.