Kunneth for two cyclic two-term complexes

The tensor total complex has groups $\mathbb{Z}$ in degree $2$, ${\mathbb{Z}}^2$ in degree $1$, and $\mathbb{Z}$ in degree $0$. In the basis $(e_1\otimes f_0,e_0\otimes f_1)$, its differentials are $d_2(k)=(-2k,2k)$ and $d_1(a,b)=2a+2b$ by the tensor differential sign rule (The tensor product of a right and a left chain complex is totalized by direct sums with the Koszul differential). Hence $\ker d_1=\{(k,-k):k\in \mathbb{Z}\}$ and $\mathrm{im}{d}_2=\{(2k,-2k):k\in \mathbb{Z}\}$, so $H_1(C\otimes D)\cong \mathbb{Z}\mathrm{/}2$.

Tensoring the specified free resolution $Q$ with $\mathbb{Z}\mathrm{/}2$ makes its differential multiplication by $2$ on $\mathbb{Z}\mathrm{/}2$, hence zero. By the fixed-resolution definition (Tor from a projective resolution of the right module), ${\mathrm{Tor}}_1^{\mathbb{Z},Q}(\mathbb{Z}\mathrm{/}2,\mathbb{Z}\mathrm{/}2)=\ker (2:\mathbb{Z}\mathrm{/}2\to \mathbb{Z}\mathrm{/}2)\cong \mathbb{Z}\mathrm{/}2$. The map $[(k,-k)]\mapsto [k{]}_2$ identifies this group with the degree-one homology from step 1.1. This is the claimed nonzero term for these fixed complexes; neither calculation selects or compares resolutions.