By Lemma \ref{lemma-perfect} every object of the source category may be represented by a complex $$ L^\bullet = \ldots \to 0 \to L^{-1} \to L^0 \to 0 \to \ldots $$ with $L^{-1}$ and $L^0$ finite projective $R$-modules. Let us temporarily call a complex of this type good. By  Derived Categories, Lemma \ref{derived-lemma-morphisms-from-projective-complex} morphisms between good complexes in the derived category are homotopy classes of maps of complexes. Thus we may work with good complexes and we can use the determinant $\det(L^\bullet) = \det(L^0) \otimes \det(L^{-1})^{\otimes -1}$ we investigated above. \medskip\noindent Let $a^\bullet : L^\bullet \to K^\bullet$ be a morphism of good complexes which is an isomorphism in $D(R)$, i.e., a quasi-isomorphism. We say that $$ \xymatrix{ L^\bullet \ar[rr]_{a^\bullet} & & K^\bullet \\ & M^\bullet \ar[lu]^{b^\bullet} \ar[ru]_{c^\bullet} } $$ is a good diagram if it commutes up to homotopy and $b^\bullet$ and $c^\bullet$ satisfy conditions (1), (2), (3) above. Whenever we have such a diagram it makes sense to define $$ \det(a^\bullet) = \det(c^\bullet)^{-1} \circ \det(b^\bullet) $$ where $\det(c^\bullet)$ and $\det(b^\bullet)$ are the isomorphisms constructed in the text above. We will show that good diagrams always exist and that the resulting map $\det(a^\bullet)$ is independent of the choice of good diagram. \medskip\noindent Existence of good diagrams for a quasi-isomorphism $a^\bullet : L^\bullet \to K^\bullet$ of good complexes. Choose a surjection $p : R^{\oplus n} \to K^{-1}$. Then we can consider the new good complex $$ M^\bullet = \ldots \to 0 \to L^{-1} \oplus R^{\oplus n} \xrightarrow{d \oplus 1} L^0 \oplus R^{\oplus n} \to 0 \to \ldots $$ with the projection map $b^\bullet : M^\bullet \to L^\bullet$ and the map $c^\bullet : M^\bullet \to K^\bullet$ using $a^{-1} \oplus p$ in degree $-1$ and using $a^0 \oplus d \circ p$ in degree $0$. The maps $b^\bullet : M^\bullet \to L^\bullet$ and $c^\bullet : M^\bullet \to K^\bullet$ satisfy conditions (1), (2), (3) above and we get a good diagram. \medskip\noindent Suppose that we have a good diagram $$ \xymatrix{ L^\bullet \ar[rr]_{\text{id}^\bullet} & & L^\bullet \\ & M^\bullet \ar[lu]^{b^\bullet} \ar[ru]_{c^\bullet} } $$ Then by Lemma \ref{lemma-homotopic-surjections} we see that $\det(c^\bullet) = \det(b^\bullet)$. Thus we see that $\det(\text{id}^\bullet) = \text{id}$ is independent of the choice of good diagram. \medskip\noindent Before we prove independence in general, we think about composition. Suppose we have quasi-isomorphisms $L_1^\bullet \to L_2^\bullet$ and $L_2^\bullet \to L_3^\bullet$ of good complexes and good diagrams $$ \vcenter{ \xymatrix{ L_1^\bullet \ar[rr] & & L_2^\bullet \\ & M_{12}^\bullet \ar[lu] \ar[ru] } } \quad\text{and}\quad \vcenter{ \xymatrix{ L_2^\bullet \ar[rr] & & L_3^\bullet \\ & M_{23}^\bullet \ar[lu] \ar[ru] } } $$ We can extend this to a diagram $$ \xymatrix{ L_1^\bullet \ar[rr] & & L_2^\bullet \ar[rr] & & L_3^\bullet \\ & M_{12}^\bullet \ar[lu] \ar[ru] & & M_{23}^\bullet \ar[lu] \ar[ru] \\ & & M_{123}^\bullet \ar[lu] \ar[ru] } $$ where $M_{123}^\bullet \to M_{12}^\bullet$ and $M_{123}^\bullet \to M_{23}^\bullet$ have properties (1), (2), (3) and the square in the diagram commutes: we can just take $M_{123}^n = M_{12}^n \times_{L_2^n} M_{23}^n$. Then Lemma \ref{lemma-compose-surjections} shows that $$ \xymatrix{ \det(L_2^\bullet) \ar[r] \ar[d] & \det(M_{23}^\bullet) \ar[d] \\ \det(M_{12}^\bullet) \ar[r] & \det(M_{123}^\bullet) } $$ commutes. A diagram chase shows that the composition $\det(L_1^\bullet) \to \det(L_2^\bullet) \to \det(L_3^\bullet)$ of the maps associated to the two good diagrams using $M_{12}^\bullet$ and $M_{23}^\bullet$ is equal to the map associated to the good diagram $$ \xymatrix{ L_1^\bullet \ar[rr] & & L_3^\bullet \\ & M_{123}^\bullet \ar[lu] \ar[ru] } $$ Thus if we can show that these maps are independent of choices, then the composition law is satisfied too and we obtain our functor. \medskip\noindent Independence. Let a quasi-isomorphism $a^\bullet : L^\bullet \to K^\bullet$ of good complexes be given. Choose an inverse quasi-isomorphism $b^\bullet : K^\bullet \to L^\bullet$. Setting $L_1^\bullet = L$, $L_2^\bullet = K^\bullet$ and $L_3^\bullet = L^\bullet$ may fix our choice of good diagram for $b^\bullet$ and consider varying good diagrams for $a^\bullet$. Then the result of the previous paragraphs is that no matter what choices, the composition always equals the identity map on $\det(L^\bullet)$. This clearly proves indepence of those choices. \medskip\noindent The statement on canonical elements follows immediately from Lemma \ref{lemma-canonical-element-well-defined} and our construction.