Let \(z \in Z_p(C)\), so \(\partial _p(z) = 0\). We must show \(\rho _p(h)(z) \in Z_p(C)\). By the equivariance hypothesis \(\partial _p \circ \rho _p(h) = \rho _{p-1}(h) \circ \partial _p\), evaluating at \(z\) gives \(\partial _p(\rho _p(h)(z)) = \rho _{p-1}(h)(\partial _p(z)) = \rho _{p-1}(h)(0) = 0\). The map \(h \mapsto \rho ^Z_p(h)\) is a group homomorphism: \(\rho ^Z_p(1) = \mathrm{id}\) follows from \(\rho _p(1) = \mathrm{id}\), and \(\rho ^Z_p(g_1 g_2) = \rho ^Z_p(g_1) \circ \rho ^Z_p(g_2)\) follows from \(\rho _p(g_1 g_2) = \rho _p(g_1) \circ \rho _p(g_2)\).

Let \(b \in B_p(C)\), so there exists \(x \in C_{p+1}\) with \(b = \partial _{p+1}(x)\). We must show \(\rho _p(h)(b) \in B_p(C)\). We claim \(\rho _p(h)(b) = \partial _{p+1}(\rho _{p+1}(h)(x))\). By the equivariance condition \(\partial _{p+1} \circ \rho _{p+1}(h) = \rho _p(h) \circ \partial _{p+1}\), evaluating at \(x\) gives \(\partial _{p+1}(\rho _{p+1}(h)(x)) = \rho _p(h)(\partial _{p+1}(x)) = \rho _p(h)(b)\). The homomorphism property follows from that of \(\rho _p\).

Let \(z\) be an element of \(C.complex.boundariesSubCycles\ p\), meaning \(z\) is a cycle and a boundary. The image \(\rho _p(h)(z)\) is a cycle by the cycles action, and it is a boundary by the boundaries action. The \(\mathbb {F}_2\)-linearity and group homomorphism property follow from those of \(\rho _p\).

We define the action using \(\operatorname {mapQ}\): for each \(h \in H\), the linear endomorphism \(\rho ^Z_p(h)\) of \(Z_p(C)\) sends \(B_p(C)\) into itself (since \(\rho ^B_p(h)\) is well-defined), so it descends to a well-defined linear map on the quotient \(H_p(C) = Z_p(C)/B_p(C)\). For the homomorphism property, \(\rho ^{H_p}(1) = \mathrm{id}\): on a quotient representative \([z]\), \([\rho _p(1)(z)] = [z]\) since \(\rho _p(1) = \mathrm{id}\). For \(g_1, g_2 \in H\), \([\rho _p(g_1 g_2)(z)] = [\rho _p(g_1)(\rho _p(g_2)(z))]\) by \(\rho _p(g_1 g_2) = \rho _p(g_1) \circ \rho _p(g_2)\), verified on each quotient element by surjectivity.

Since \(|H|\) is odd, write \(|H| = 2k + 1\) for some \(k \in \mathbb {N}\). Then in \(\mathbb {F}_2\): \((|H| : \mathbb {F}_2) = (2k + 1 : \mathbb {F}_2) = 2k + 1 = 0 \cdot k + 1 = 1,\) using \((2 : \mathbb {F}_2) = 0\) (verified by computation). Hence \((|H| : \mathbb {F}_2) = 1\), which is invertible (with \(\operatorname {Invertible} 1\) given by the standard instance for units).

Take \(H = \mathbf{1}\) (the trivial group). Then \(|H| = 1\), which is odd. Let \(C = D\) be the trivial \(H\)-equivariant chain complex given by the single-object chain complex concentrated at degree \(0\) with value \(\mathbb {F}_2\), equipped with the trivial \(H\)-action (the only possible action since \(H\) is trivial). This is equivariant vacuously. Thus the witness is \((H, C, D)\) with \(\operatorname {Odd}(|\mathbf{1}|)\) witnessed by \(1 = 2 \cdot 0 + 1\).

The zero element \(0\) belongs to every module, so \(H_p(C) \otimes _H H_{n-p}(D)\) contains at least \(0\).

The zero element \(0\) belongs to every direct sum of modules, so the balanced Künneth sum contains at least \(0\).