The Stacks project

Proof. We will only prove (1). Let $U \subset X$ be the open subspace used in the construction of $Z'$, i.e., an open such that $Z$ is identified with a closed subspace of $U$ cut out by the quasi-coherent sheaf of ideals $\mathcal{I}$. Since $|T| = |T'|$ we see that $|b|(|T'|) \subset |U|$. Hence we can think of $b$ as a morphism into $U$, see Properties of Spaces, Lemma 66.4.9. Let $\mathcal{J} \subset \mathcal{O}_{T'}$ be the square zero quasi-coherent sheaf of ideals cutting out $T$. By the commutativity of the diagram we have $b|_ T = i \circ a$ where $i : Z \to U$ is the closed immersion. We conclude that $b^\sharp (b^{-1}\mathcal{I}) \subset \mathcal{J}$ by Morphisms of Spaces, Lemma 67.13.1. As $T'$ is a first order thickening of $T$ we see that $\mathcal{J}^2 = 0$ hence $b^\sharp (b^{-1}(\mathcal{I}^2)) = 0$. By Morphisms of Spaces, Lemma 67.13.1 this implies that $b$ factors through $Z'$. Letting $a' : T' \to Z'$ be this factorization we win. $\square$