Lemma: Cross-entry null relations

$\ker \mathsf{Rec}\supseteq {⨁}_e\ker {\mathsf{Rec}}_e$, with equality if and only if the sum ${\sum }_e{\mathcal{K}}_e\subseteq {\mathcal{D}}^{\prime }(W)$ is direct. If ${\mathcal{K}}_e\cap {\mathcal{K}}_{e^{\prime }}\ne 0$ for some $e\ne e^{\prime }$, then $\ker \mathsf{Rec}$ contains elements $(p_e,-p_{e^{\prime }})$ with ${\mathsf{Rec}}_e{p}_e={\mathsf{Rec}}_{e^{\prime }}{p}_{e^{\prime }}\ne 0$, and by (eq:descent) every joint map must take equal values on the $e$- and $e^{\prime }$-presentations of one and the same kernel.

Proof. If ${\sum }_e{\mathsf{Rec}}_e{p}_e=0$ and the sum of images is direct, each ${\mathsf{Rec}}_e{p}_e=0$; conversely a nontrivial relation ${\sum }_e{k}_e=0$, $k_e\in {\mathcal{K}}_e$ not all zero, lifts to an element of $\ker \mathsf{Rec}$ outside $⨁\ker {\mathsf{Rec}}_e$. The last sentence is (eq:descent) applied to $(p_e,-p_{e^{\prime }})$.