**Proof.**
Let us prove (1). Set $\psi = 0$. Then we may, with notation as above Definition 42.68.13, identify $K_\varphi = I_\psi = 0$, $I_\varphi = K_\psi = M$. With these identifications, the map

\[ \gamma _\varphi : \kappa \otimes \det \nolimits _\kappa (M) = \det \nolimits _\kappa (K_\varphi ) \otimes \det \nolimits _\kappa (I_\varphi ) \longrightarrow \det \nolimits _\kappa (M) \]

is identified with $\det _\kappa (\varphi ^{-1})$. On the other hand the map $\gamma _\psi $ is identified with the identity map. Hence $\gamma _\psi \circ \sigma \circ \gamma _\varphi ^{-1}$ is equal to $\det _\kappa (\varphi )$ in this case. Whence the result. We omit the proof of (2).
$\square$

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