Lemma 15.6.6. In Situation 15.6.1 let $(N_1, M'_1, \varphi _1) \to (N_2, M'_2, \varphi _2)$ be a morphism of $\text{Mod}_ B \times _{\text{Mod}_ A} \text{Mod}_{A'}$ with $N_1 \to N_2$ and $M'_1 \to M'_2$ surjective. Then

$N_1 \times _{\varphi _1, M_1} M'_1 \to N_2 \times _{\varphi _2, M_2} M'_2$

where $M_1 = M'_1/IM'_1$ and $M_2 = M'_2/IM'_2$ is surjective.

Proof. Pick $(x_2, y_2) \in N_2 \times _{\varphi _2, M_2} M'_2$. Choose $x_1 \in N_1$ mapping to $x_2$. Since $M'_1 \to M_1$ is surjective we can find $y_1 \in M'_1$ mapping to $\varphi _1(x_1)$. Then $(x_1, y_1)$ maps to $(x_2, y'_2)$ in $N_2 \times _{\varphi _2, M_2} M'_2$. Thus it suffices to show that elements of the form $(0, y_2)$ are in the image of the map. Here we see that $y_2 \in IM'_2$. Write $y_2 = \sum t_ i y_{2, i}$ with $t_ i \in I$. Choose $y_{1, i} \in M'_1$ mapping to $y_{2, i}$. Then $y_1 = \sum t_ iy_{1, i} \in IM'_1$ and the element $(0, y_1)$ does the job. $\square$

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