The Stacks project

Proof. We first construct $Y'$ as a ringed space. Namely, as topological space we take $Y' = Y$. Denote $f' : X' \to Y'$ the map of topological spaces which equals $f$. As structure sheaf $\mathcal{O}_{Y'}$ we take the right hand side of the equation of the lemma. To see that $Y'$ is a scheme, we have to show that any point has an affine neighbourhood. Since the formation of the fibre product of sheaves commutes with restricting to opens, we may assume $Y$ is affine. Then $X$ is affine (as $f$ is affine) and $X'$ is affine as well (see Lemma 37.2.3). Say $Y \leftarrow X \rightarrow X'$ corresponds to $B \rightarrow A \leftarrow A'$. Set $B' = B \times _ A A'$; this is the global sections of $\mathcal{O}_{Y'}$. As $A' \to A$ is surjective with locally nilpotent kernel we see that $B' \to B$ is surjective with locally nilpotent kernel. Hence $\mathop{\mathrm{Spec}}(B') = \mathop{\mathrm{Spec}}(B)$ (as topological spaces). We claim that $Y' = \mathop{\mathrm{Spec}}(B')$. To see this we will show for $g' \in B'$ with image $g \in B$ that $\mathcal{O}_{Y'}(D(g)) = B'_{g'}$. Namely, by More on Algebra, Lemma 15.5.3 we see that

where $h \in A$, $h' \in A'$ are the images of $g'$. Since $B_ g$, resp. $A_ h$, resp. $A'_{h'}$ is equal to $\mathcal{O}_ Y(D(g))$, resp. $f_*\mathcal{O}_ X(D(g))$, resp. $f'_*\mathcal{O}_{X'}(D(g))$ the claim follows.

It remains to show that $Y'$ is the pushout. The discussion above shows the scheme $Y'$ has an affine open covering $Y' = \bigcup W'_ i$ such that the corresponding opens $U'_ i \subset X'$, $W_ i \subset Y$, and $U_ i \subset X$ are affine open. Moreover, if $A'_ i$, $B_ i$, $A_ i$ are the rings corresponding to $U'_ i$, $W_ i$, $U_ i$, then $W'_ i$ corresponds to $B_ i \times _{A_ i} A'_ i$. Thus we can apply Lemmas 37.14.1 and 37.14.2 to conclude our construction is a pushout in the category of schemes. $\square$