We verify both conditions for complementarity.

Disjointness (\(H_1^L \cap H_1^G = \{ 0\} \)): We use the submodule disjointness criterion. Let \(x \in H_1^L \cap H_1^G\). Then by definition of \(H_1^L = \operatorname {im}(\operatorname {embH})\) and \(H_1^G = \operatorname {im}(\operatorname {embV})\), there exist \(a \in H_1^h\) and \(b \in H_1^v\) such that \(x = \operatorname {embH}(a) = \operatorname {embV}(b)\).

Let \(\kappa \) denote the Künneth isomorphism. Applying \(\kappa \) to both sides, we obtain

Since \(\operatorname {embH} = \kappa ^{-1} \circ \operatorname {incH}\) and \(\operatorname {embV} = \kappa ^{-1} \circ \operatorname {incV}\), applying \(\kappa \) gives \(\kappa \circ \kappa ^{-1} = \operatorname {id}\), so:

Now \(\operatorname {incH}\) includes into the direct summand at index \(p = 1\), and \(\operatorname {incV}\) includes into the summand at \(p = 0\). Applying the component projection at \(p = 1\) to both sides:

since \(\pi _1(\operatorname {incV}(b)) = 0\) as \(\operatorname {incV}\) maps into the \(p = 0\) summand. Thus \(a = 0\), and by simplification \(x = \operatorname {embH}(0) = 0\).

Codisjointness (\(H_1^L + H_1^G = H_1\)): We show every \(x \in H_1\) can be written as a sum of an element of \(H_1^L\) and an element of \(H_1^G\). Let \(z = \kappa (x)\) be the image of \(x\) under the Künneth isomorphism. Since the balanced Künneth sum at degree 1 decomposes over the summands at \(p = 0\) and \(p = 1\) (all other summands are trivial, as the Tanner and cycle complexes have PUnit modules outside degrees \(\{ 0,1\} \)), we have

where \(z_h = \pi _1(z) \in H_1^h\) and \(z_v = \pi _0(z) \in H_1^v\) are the components at \(p = 1\) and \(p = 0\) respectively, and \(\iota _p = \operatorname {directSumInc}(-,p)\) are the canonical inclusions.

Set \(x_h = \operatorname {embH}(z_h) \in H_1^L\) and \(x_v = \operatorname {embV}(z_v) \in H_1^G\). We claim \(x = x_h + x_v\). Indeed, applying \(\kappa \) (which is injective):

By injectivity of \(\kappa \), we conclude \(x = x_h + x_v\), completing the proof.

Since over \(\mathbb {F}_2\) the cohomology summands \(\operatorname {coH}1h\) and \(\operatorname {coH}1v\) defined in Def. 27 coincide definitionally with the homology summands \(H_1^h\) and \(H_1^v\) respectively, the cohomology logical and gauge submodules \(\operatorname {coHL}\) and \(\operatorname {coHG}\) are definitionally equal to \(H_1^L\) and \(H_1^G\). The result therefore follows immediately from the homology complementarity theorem \(\operatorname {hcompl}\).

We show that \(H_1^L = \operatorname {im}(\operatorname {embH}) = \{ 0\} \) by proving every element in the image is zero.

Let \(x \in H_1^L\), so there exists \(a \in H_1^h\) with \(x = \operatorname {embH}(a)\). We claim \(a = 0\).

First, note that the cycles at degree 1 of the quotient Tanner complex equal the kernel of the quotient differential: \(Z_1(\mathcal{C}(X/H,\Lambda )) = \ker (\partial _{\mathrm{quot}})\). Since \(\ker (\partial _{\mathrm{quot}}) = \{ 0\} \) by hypothesis, the cycles submodule is trivial, and consequently the quotient Tanner homology \(H_1(\mathcal{C}(X/H, \Lambda ))\) is trivial (every element equals zero, since the homology is a quotient of a trivial submodule).

By the iota/pi isomorphism of Def. 28 (\(\ell \) odd, \(|H|\) odd), we have \(\operatorname {iota} \circ \operatorname {pi} = \operatorname {id}\) on \(H_1^h\). Applying this identity to \(a\):

Since \(H_1(\mathcal{C}(X/H,\Lambda ))\) is trivial, \(\operatorname {pi}(a) = 0\), and therefore

Thus \(x = \operatorname {embH}(0) = 0\), completing the proof that \(H_1^L = \{ 0\} \).

We show \(H_1^G = \operatorname {im}(\operatorname {embV}) = \{ 0\} \) by proving every element in the image is zero.

Let \(x \in H_1^G\), so there exists \(a \in H_1^v\) with \(x = \operatorname {embV}(a)\). We claim \(a = 0\).

Recall that \(H_1^v\) is the balanced Künneth summand at \(p = 0\):

Since \(H_0(\mathcal{C}(X,\Lambda ))\) is subsingleton by hypothesis, the tensor product \(H_0(\mathcal{C}(X,\Lambda )) \otimes _{\mathbb {F}_2} H_1(\mathcal{C}_\ell )\) is also subsingleton. Indeed, any element of the tensor product can be written as a sum of pure tensors \(m \otimes n\); since \(m \in H_0(\mathcal{C}(X,\Lambda ))\) is forced to be zero (the module is subsingleton), each pure tensor equals \(0 \otimes n = 0\), so the entire element is zero.

Since the coinvariants module is a quotient of a subsingleton module, it is itself subsingleton. Therefore \(H_1^v\) is subsingleton, which forces \(a = 0\).

Hence \(x = \operatorname {embV}(0) = 0\), so \(H_1^G = \{ 0\} \).

Since \(H_1^L\) is a submodule of \(H_1\), it contains the zero element. Therefore its carrier is nonempty, witnessed by \(0 \in H_1^L\).

Since \(H_1^G\) is a submodule of \(H_1\), it contains the zero element. Therefore its carrier is nonempty, witnessed by \(0 \in H_1^G\).