It is clear that the assumptions imply part (1) of the lemma. \medskip\noindent 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. \medskip\noindent 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. \medskip\noindent 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 \begin{enumerate} \item an admissible sequence $x_1, \ldots, x_r \in K_\alpha$ generating $K_\alpha$ \item an admissible sequence $y_1, \ldots, y_s \in K_\beta$ generating $K_\beta$, \item an admissible sequence $z_1, \ldots, z_t \in K_\gamma$ generating $K_\gamma$, \item elements $\tilde x_i \in M$ such that $\beta\gamma\tilde x_i = x_i$, \item elements $\tilde y_i \in M$ such that $\alpha\gamma\tilde y_i = y_i$, \item elements $\tilde z_i \in M$ such that $\beta\alpha\tilde z_i = z_i$. \end{enumerate} 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 \ref{remark-more-elementary} 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.