The Stacks project

Proof. By More on Algebra, Lemma 15.6.2 we have $T' = T \amalg _ S S'$ as topological spaces, i.e., the diagram is a pushout in the category of topological spaces. Next, consider the map

where $g = i' \circ f = f' \circ i$. We claim this map is an isomorphism of sheaves of rings. Namely, we can view both sides as quasi-coherent $\mathcal{O}_{T'}$-modules (use Schemes, Lemmas 26.24.1 for the right hand side) and the map is $\mathcal{O}_{T'}$-linear. Thus it suffices to show the map is an isomorphism on the level of global sections (Schemes, Lemma 26.7.5). On global sections we recover the identification $B' \to B \times _ A A'$ from statement of the lemma (this is how we chose $B'$).

Let $X$ be a scheme. Suppose we are given morphisms of schemes $m' : S' \to X$ and $n : T \to X$ such that $m' \circ i = n \circ f$ (call this $m$). We get a unique map of topological spaces $n' : T' \to X$ compatible with $m'$ and $n$ as $T' = T \amalg _ S S'$ (see above). By the description of $\mathcal{O}_{T'}$ in the previous paragraph we obtain a unique homomorphism of sheaves of rings

given by $(m')^\sharp $ and $n^\sharp $. Thus $(n', (n')^\sharp )$ is the unique morphism of ringed spaces $T' \to X$ compatible with $m'$ and $n$. To finish the proof it suffices to show that $n'$ is a morphism of schemes, i.e., a morphism of locally ringed spaces.

Let $t' \in T'$ with image $x \in X$. We have to show that $\mathcal{O}_{X, x} \to \mathcal{O}_{T', t'}$ is local. If $t' \not\in T$, then $t'$ is the image of a unique point $s' \in S'$ and $\mathcal{O}_{T', t'} = \mathcal{O}_{S', s'}$. Namely, $S' \setminus S \to T' \setminus T$ is an isomorphism of schemes as $B' \to A'$ induces an isomorphism $\mathop{\mathrm{Ker}}(B' \to B) = \mathop{\mathrm{Ker}}(A' \to A)$. If $t'$ is the image of $t \in T$, then we know that the composition $\mathcal{O}_{X, x} \to \mathcal{O}_{T', t'} \to \mathcal{O}_{T, t}$ is local and we conclude also. $\square$

Proof. Namely, for a scheme $T$ a morphism $W \to T$ is the same thing as collection of morphism $W_ a \to T$, $a \in A$ which agree on the overlaps $W_ a \times _ W W_ b$, see Descent, Lemma 35.13.7. $\square$

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. $\square$

Proof. Let $(V, U', \varphi )$ be an object of the fibre product category. Set $U = U' \times _{X'} X$. Note that $U \to U'$ is a thickening. Since $\varphi : V \times _ Y X \to U' \times _{X'} X = U$ is an isomorphism we have a morphism $U \to V$ over $X \to Y$ which identifies $U$ with the fibre product $X \times _ Y V$. In particular $U \to V$ is affine, see Morphisms, Lemma 29.11.8. Hence we can apply Lemma 37.14.3 to get a pushout $V' = V \amalg _ U U'$. Denote $V' \to Y'$ the morphism we obtain in virtue of the fact that $V'$ is a pushout and because we are given morphisms $V \to Y$ and $U' \to X'$ agreeing on $U$ as morphisms into $Y'$. Setting $G(V, U', \varphi ) = V'$ gives the functor $G$.

Let us prove that $G$ is a left adjoint to $F$. Let $Z$ be a scheme over $Y'$. We have to show that

where the morphism sets are taking in their respective categories. Let $g' : V' \to Z$ be a morphism. Denote $\tilde g$, resp. $\tilde f'$ the composition of $g'$ with the morphism $V \to V'$, resp. $U' \to V'$. Base change $\tilde g$, resp. $\tilde f'$ by $Y \to Y'$, resp. $X' \to Y'$ to get a morphism $g : V \to Z \times _{Y'} Y$, resp. $f' : U' \to Z \times _{Y'} X'$. Then $(g, f')$ is an element of the right hand side of the equation above (details omitted). Conversely, suppose that $(g, f') : (V, U', \varphi ) \to F(Z)$ is an element of the right hand side. We may consider the composition $\tilde g : V \to Z$, resp. $\tilde f' : U' \to Z$ of $g$, resp. $f$ by $Z \times _{Y'} X' \to Z$, resp. $Z \times _{Y'} Y \to Z$. Then $\tilde g$ and $\tilde f'$ agree as morphism from $U$ to $Z$. By the universal property of pushout, we obtain a morphism $g' : V' \to Z$, i.e., an element of the left hand side. We omit the verification that these constructions are mutually inverse.

To prove that $F \circ G$ is isomorphic to the identity we have to show that the adjunction mapping $(V, U', \varphi ) \to F(G(V, U', \varphi ))$ is an isomorphism. To do this we may work affine locally. Say $X = \mathop{\mathrm{Spec}}(A)$, $X' = \mathop{\mathrm{Spec}}(A')$, and $Y = \mathop{\mathrm{Spec}}(B)$. Then $A' \to A$ and $B \to A$ are ring maps as in More on Algebra, Lemma 15.6.4 and $Y' = \mathop{\mathrm{Spec}}(B')$ with $B' = B \times _ A A'$. Next, suppose that $V = \mathop{\mathrm{Spec}}(D)$, $U' = \mathop{\mathrm{Spec}}(C')$ and $\varphi $ is given by an $A$-algebra isomorphism $D \otimes _ B A \to C' \otimes _{A'} A = C'/IC'$. Set $D' = D \times _{C'/IC'} C'$. In this case the statement we have to prove is that $D' \otimes _{B'} B \cong D$ and $D' \otimes _{B'} A' \cong C'$. This is a special case of More on Algebra, Lemma 15.6.4. $\square$

Proof. A quasi-inverse functor assigns to the triple $(\mathcal{G}, \mathcal{F}', \varphi )$ the fibre product

where $\mathcal{F} = (U \to U')^*\mathcal{F}'$. This works, because on affines we recover the equivalence of More on Algebra, Lemma 15.7.5. Some details omitted.

Parts (a) and (b) follow from More on Algebra, Lemmas 15.7.4 and 15.7.6. $\square$

Proof. Looking over affine pieces the assertions of this lemma are equivalent to the corresponding assertions of More on Algebra, Lemma 15.7.7. $\square$