Proof.
By Lemma 15.74.2 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 13.19.8 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.
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
\[ \det (a^\bullet ) = \det (c^\bullet ) \circ \det (b^\bullet )^{-1} \]
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.
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.
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 15.122.2 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.
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 15.122.3 shows that
\[ \xymatrix{ \det (L_2^\bullet ) & \det (M_{23}^\bullet ) \ar[l] \\ \det (M_{12}^\bullet ) \ar[u] & \det (M_{123}^\bullet ) \ar[l] \ar[u] } \]
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.
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.
The statement on canonical elements follows immediately from Lemma 15.122.1 and our construction.
$\square$
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