I am trying to understand the de Rham double complex associated to a Lie groupoid, and I am having trouble proving a fundamental property of the total differential, i.e., that it squares to zero.
Given a Lie groupoid, we can form a simplicial manifold $\Gamma_{\bullet}$ by taking $\Gamma_p$ to be the set of $p$ composable arrows in the groupoid.
Then, we define the de Rham double complex in degree $p$ to be the direct sum of differential form complexes on the spaces $\Gamma_p$. The differentials are given by the de Rham exterior derivative $d$, which increases the degree of the forms, and the alternating sum of pullbacks under the face maps of the simplicial manifold, denoted by $\partial$.
The total differential on this double complex is then defined as a sum of these two maps: $\delta = (-1)^p d + \partial$. The crucial property I want to show is that this total differential squares to zero, i.e., $\delta^2 = 0$.
I already know that $d^2$, how can then show that $\partial^2=0$ and $(-1)^p\partial d+(-1)^p d\partial=0$?
Could anyone provide a detailed proof of this, or point me to a source where this is proven? Any help would be appreciated. Thanks