Fix direction det operator

Thanks to Tuomas Tajakka
https://stacks.math.columbia.edu/tag/0FJI#comment-5958

more-algebra.tex

@@ -34276,43 +34276,38 @@ \section{Determinants of complexes}
 \item $K^n$, $L^n$ are finite projective $R$-modules, nonzero only for
 $n \in \{-1, 0\}$.
 \end{enumerate}
-Then we construct an isomorphism
+In this situation we will construct an isomorphism
 $$
 \det(a^\bullet) : \det(K^\bullet) \longrightarrow \det(L^\bullet)
 $$
-as follows. Using the exact sequences
+Using the exact sequences $0 \to \Ker(a^i) \to K^i \to L^i \to 0$
+we obtain isomorphisms
 $$
-0 \to \Ker(a^i) \to K^i \to L^i \to 0
+\gamma^i : \det(\Ker(a^i)) \otimes \det(L^i) \to \det(K^i)
 $$
-we obtain isomorphisms
-$\gamma^i : \det(\Ker(a^i)) \otimes \det(L^i) \to \det(K^i)$
-by Lemma \ref{lemma-det-ses}. Since $a^\bullet$ is a quasi-isomorphism
-the complex $\Ker(a^\bullet)$ is acyclic, has rank $0$, and the
-canonical section defines a trivialization of the
-invertible $R$-module $\det(\Ker(a^\bullet)) =
-\det(\Ker(a^0)) \otimes \det(\Ker(a^{-1}))^{\otimes -1}$, see above.
-Then we define $\det(a^\bullet)$ as the map
-\begin{align*}
-\det(L^\bullet)
-& =
-\det(L^0) \otimes_R \det(L^{-1})^{\otimes -1} \\
-& \to
-\det(\Ker(a^0)) \otimes_R \det(L^0) \otimes_R
-\left(\det(\Ker(a^{-1}) \otimes_R \det(L^{-1})\right)^{\otimes -1} \\
-& \to
-\det(K^0) \otimes_R \det(K^{-1})^{\otimes -1} \\
-& =
-\det(K^\bullet)
-\end{align*}
-where the first arrow uses the trivialization of $\det(\Ker(a^\bullet))$
-and the second arrow uses the isomorphisms $\gamma^i$.
+for $i = -1, 0$ by Lemma \ref{lemma-det-ses}. Since $a^\bullet$
+is a quasi-isomorphism the complex $\Ker(a^\bullet)$ is acyclic
+and has rank $0$. Hence the canonical element $\delta(\Ker(a^\bullet))$
+is a trivialization of the invertible $R$-module
+$\det(\Ker(a^\bullet))$, see above. We define
+$\det(a^\bullet) : \det(K^\bullet) \to \det(L^\bullet)$ as the
+unique isomorphism such that the diagram
+$$
+\xymatrix{
+\det(K^\bullet) \ar[rr]_{\det(a^\bullet)} \ar[dr]_{\delta(\Ker(a^\bullet))} & &
+\det(L^\bullet) \\
+& \det(K^\bullet) \otimes \det(\Ker(a^\bullet))
+\ar[ru]_{\gamma^0 \otimes (\gamma^{-1})^{\otimes -1}}
+}
+$$
+commutes.
 
 \begin{lemma}
 \label{lemma-canonical-element-well-defined}
 Let $R$ be a ring. Let $a^\bullet : K^\bullet \to L^\bullet$ be a map of
 complexes of $R$-modules satisfying (1), (2), (3) above. If $L^\bullet$
 has rank $0$, then $\det(a^\bullet)$ maps the
-canonical element $\delta(L^\bullet)$ to $\delta(K^\bullet)$.
+canonical element $\delta(K^\bullet)$ to $\delta(L^\bullet)$.
 \end{lemma}
 
 \begin{proof}
@@ -34345,7 +34340,7 @@ \section{Determinants of complexes}
 Let $h : K^0 \to L^{-1}$ be a map such that
 $b^0 = a^0 + d \circ h$ and $b^{-1} = a^{-1} + h \circ d$ are surjective.
 Then $\det(a^\bullet) = \det(b^\bullet)$ as maps
-$\det(L^\bullet) \to \det(K^\bullet)$.
+$\det(K^\bullet) \to \det(L^\bullet)$.
 \end{lemma}
 
 \begin{proof}
@@ -34380,7 +34375,7 @@ \section{Determinants of complexes}
 }
 $$
 commutes. Since $\det(c^\bullet)$ maps the canonical trivialization
-of $\det(\Ker(a^\bullet))$ to the same for $b^\bullet$
+of $\det(\Ker(a^\bullet))$ to the canonical trivializatio of $\Ker(b^\bullet)$
 (Lemma \ref{lemma-canonical-element-well-defined})
 we see that we conclude if (and only if)
 $$
@@ -34435,7 +34430,7 @@ \section{Determinants of complexes}
 Let $R$ be a ring. Let $a^\bullet : K^\bullet \to L^\bullet$
 and $b^\bullet : L^\bullet \to M^\bullet$ be maps of
 complexes of $R$-modules satisfying (1), (2), (3) above.
-Then we have $\det(a^\bullet) \circ \det(b^\bullet) = 
+Then we have $\det(b^\bullet) \circ \det(a^\bullet) = 
 \det(b^\bullet \circ a^\bullet)$ as maps
 $\det(M^\bullet) \to \det(K^\bullet)$.
 \end{lemma}
@@ -34500,7 +34495,7 @@ \section{Determinants of complexes}
 and $c^\bullet$ satisfy conditions (1), (2), (3) above. Whenever
 we have such a diagram it makes sense to define
 $$
-\det(a^\bullet) = \det(c^\bullet)^{-1} \circ \det(b^\bullet)
+\det(a^\bullet) = \det(c^\bullet) \circ \det(b^\bullet)^{-1}
 $$
 where $\det(c^\bullet)$ and $\det(b^\bullet)$ are the isomorphisms
 constructed in the text above. We will show that good diagrams always
@@ -34576,14 +34571,13 @@ \section{Determinants of complexes}
 $M_{123}^\bullet \to M_{23}^\bullet$ 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
+Then Lemma \ref{lemma-compose-surjections} shows that
 $$
 \xymatrix{
-\det(L_2^\bullet) \ar[r] \ar[d] &
-\det(M_{23}^\bullet) \ar[d] \\
-\det(M_{12}^\bullet) \ar[r] &
-\det(M_{123}^\bullet)
+\det(L_2^\bullet) &
+\det(M_{23}^\bullet) \ar[l] \\
+\det(M_{12}^\bullet) \ar[u] &
+\det(M_{123}^\bullet) \ar[l] \ar[u]
 }
 $$
 commutes. A diagram chase shows that the composition