Theorem: Free exactness and exact tangents

For a differentiable solution $A_{\xi }c(\xi )+d_{\xi }=0$ in the fixed identified presentation,

At fixed $\xi$, put $w_{\xi }=-d_{\xi }^{\prime }-A_{\xi }^{\prime }c(\xi )$. An exact tangent $z=Su$ satisfying differentiated cancellation exists iff

Every such tangent is $B_0$-closed.

Proof. Apply Theorem thm:rank to $S$. Differentiate cancellation and substitute $c^{\prime }=Su$. Finally $B_{0}Su=0$.