The diagonal of a finite étale algebra contracts its positive Hochschild cochains

The open and closed diagonal in $\mathrm{Spec}(D{\otimes }_{A}D)$ corresponds to an idempotent $e$ on whose summand the multiplication map $\mu :D{\otimes }_{A}D\to D$ is an isomorphism, while the complementary summand is its kernel. Hence $\mu (e)=1$. For $a\in D$, $a\otimes 1-1\otimes a$ is in that kernel and therefore annihilates $e$. Expressing the tensor $e$ as a finite sum gives the asserted identities.

Agreement of the two $A$-actions makes $\delta$ preserve $A$-multilinearity and makes the tensor formula for $h$ well-defined. Expand $\delta (hc)+h(\delta c)$ using the displayed differential. The initial term in $h\delta c$ is ${\sum }_i{x}_i{y}_{i}c(a_1,\dots ,a_n)=c(a_1,\dots ,a_n)$. Every term combining two adjacent inputs cancels with its counterpart of opposite sign in $\delta hc$, and the last right-action terms cancel as well. The remaining pair is $a_1{\sum }_i{x}_{i}c(y_i,a_2,\dots )-{\sum }_i{x}_{i}c(y_i{a}_1,a_2,\dots )$, which is zero by $(a_1\otimes 1)e=(1\otimes a_1)e$ and multilinearity. Thus $\delta h+h\delta =\mathrm{id}$ for $n\ge 1$. For a cocycle $c$, this gives $c=\delta (hc)$. An input equal to $1$ after the inserted first input still makes each summand zero, proving the normalization assertion.