Lemma 76.12.2. Let $i : Z \to X$ be an immersion of algebraic spaces. The first order infinitesimal neighbourhood $Z'$ of $Z$ in $X$ has the following universal property: Given any commutative diagram

$\xymatrix{ Z \ar[d]_ i & T \ar[l]^ a \ar[d] \\ X & T' \ar[l]_ b }$

where $T \subset T'$ is a first order thickening over $X$, there exists a unique morphism $(a', a) : (T \subset T') \to (Z \subset Z')$ of thickenings over $X$.

Proof. 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. $\square$

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