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. $\square$