Lemma 15.7.2. In Situation 15.7.1 the functor (15.7.1.1) has a right adjoint, namely the functor

$F : (N, M', \varphi ) \longmapsto N \times _{\varphi , M} M'$

where $M = M'/IM'$. Moreover, the composition of $F$ with (15.7.1.1) is the identity functor on $\text{Mod}_ D \times _{\text{Mod}_ C} \text{Mod}_{C'}$. In other words, setting $N' = N \times _{\varphi , M} M'$ we have $N' \otimes _{D'} D = N$ and $N' \otimes _{D'} C' = M'$.

Proof. The adjointness statement follows from the more general Lemma 15.5.4. The final assertion follows from the corresponding assertion of Lemma 15.6.4 because $N' \otimes _{D'} D = N' \otimes _{D'} D' \otimes _{B'} B = N' \otimes _{B'} B$ and $N' \otimes _{D'} C' = N' \otimes _{D'} D' \otimes _{B'} A' = N' \otimes _{B'} A'$. $\square$

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