Proof.
It is clear that the assumptions imply part (1) of the lemma.
To see part (1) note that the assumptions imply that I_{\gamma \alpha } = I_{\alpha \gamma }, and similarly for kernels and any other pair of morphisms. Moreover, we see that I_{\gamma \beta } =I_{\beta \gamma } = K_\alpha \subset I_\gamma and similarly for any other pair. In particular we get a short exact sequence
0 \to I_{\beta \gamma } \to I_\gamma \xrightarrow {\alpha } I_{\alpha \gamma } \to 0
and similarly we get a short exact sequence
0 \to I_{\alpha \gamma } \to I_\gamma \xrightarrow {\beta } I_{\beta \gamma } \to 0.
This proves (I_\gamma , \alpha , \beta ) is an exact (2, 1)-periodic complex. Hence part (2) of the lemma holds.
To see that \alpha , \gamma give well defined endomorphisms of M/K_\beta we have to check that \alpha (K_\beta ) \subset K_\beta and \gamma (K_\beta ) \subset K_\beta . This is true because \alpha (K_\beta ) = \alpha (I_{\gamma \alpha }) = I_{\alpha \gamma \alpha } \subset I_{\alpha \gamma } = K_\beta , and similarly in the other case. The kernel of the map \alpha : M/K_\beta \to M/K_\beta is K_{\beta \alpha }/K_\beta = I_\gamma /K_\beta . Similarly, the kernel of \gamma : M/K_\beta \to M/K_\beta is equal to I_\alpha /K_\beta . Hence we conclude that (3) holds.
We introduce r = \text{length}_ R(K_\alpha ), s = \text{length}_ R(K_\beta ) and t = \text{length}_ R(K_\gamma ). By the exact sequences above and our hypotheses we have \text{length}_ R(I_\alpha ) = s + t, \text{length}_ R(I_\beta ) = r + t, \text{length}_ R(I_\gamma ) = r + s, and \text{length}(M) = r + s + t. Choose
an admissible sequence x_1, \ldots , x_ r \in K_\alpha generating K_\alpha
an admissible sequence y_1, \ldots , y_ s \in K_\beta generating K_\beta ,
an admissible sequence z_1, \ldots , z_ t \in K_\gamma generating K_\gamma ,
elements \tilde x_ i \in M such that \beta \gamma \tilde x_ i = x_ i,
elements \tilde y_ i \in M such that \alpha \gamma \tilde y_ i = y_ i,
elements \tilde z_ i \in M such that \beta \alpha \tilde z_ i = z_ i.
With these choices the sequence y_1, \ldots , y_ s, \alpha \tilde z_1, \ldots , \alpha \tilde z_ t is an admissible sequence in I_\alpha generating it. Hence, by Remark 42.68.14 the determinant D = \det _\kappa (M, \alpha , \beta \gamma ) is the unique element of \kappa ^* such that
\begin{align*} [y_1, \ldots , y_ s, \alpha \tilde z_1, \ldots , \alpha \tilde z_ s, \tilde x_1, \ldots , \tilde x_ r] \\ = (-1)^{r(s + t)} D [x_1, \ldots , x_ r, \gamma \tilde y_1, \ldots , \gamma \tilde y_ s, \tilde z_1, \ldots , \tilde z_ t] \end{align*}
By the same remark, we see that D_1 = \det _\kappa (M/K_\beta , \alpha , \gamma ) is characterized by
[y_1, \ldots , y_ s, \alpha \tilde z_1, \ldots , \alpha \tilde z_ t, \tilde x_1, \ldots , \tilde x_ r] = (-1)^{rt} D_1 [y_1, \ldots , y_ s, \gamma \tilde x_1, \ldots , \gamma \tilde x_ r, \tilde z_1, \ldots , \tilde z_ t]
By the same remark, we see that D_2 = \det _\kappa (I_\gamma , \alpha , \beta ) is characterized by
[y_1, \ldots , y_ s, \gamma \tilde x_1, \ldots , \gamma \tilde x_ r, \tilde z_1, \ldots , \tilde z_ t] = (-1)^{rs} D_2 [x_1, \ldots , x_ r, \gamma \tilde y_1, \ldots , \gamma \tilde y_ s, \tilde z_1, \ldots , \tilde z_ t]
Combining the formulas above we see that D = D_1 D_2 as desired.
\square
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