The Stacks project

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. $\square$