I am studying this papper

By Daniel Dugger. In Section 2 Dugger proves that if $I$ as a perfect ideal such that $\operatorname{grade}(I,R)=d$ and $ \textbf{x}=x_1,\dots ,x_d$ is a regular sequence contained in $I$ such that $(x_1,\dots ,x_d) \neq I$ and $J=(\textbf{x}):I$, then $J$ is perfect and $\operatorname{grade}(J,R)=d$.

During the proof he consider $0 \to F_d \to F_{d-1} \to \cdots \to F_0 \to R/I $ be a minimal free resolutions of $R/I$ and $K_{\bullet}$ the Koszul complex of $x_1,\dots,x_d$. By comparison theorem the projection $R/(\textbf{x}) \to R/I$ lifts to a map os complexes $\pi:K_{\bullet} \to F_{\bullet}$. He says that a free resolution for $R/J$ can be obtained from the following diagram by dualizing and then taking the mapping cone.

Can someone help-me with this argument?