During this proof we will say $Z \subset Z'$ is a universal first order thickening of $Z$ over $X$ if it satisfies the condition of the lemma. We will construct the universal first order thickening $Z \subset Z'$ over $X$ by glueing, starting with the affine case which is Algebra, Lemma \ref{algebra-lemma-universal-thickening}. We begin with some general remarks. \medskip\noindent If a universal first order thickening of $Z$ over $X$ exists, then it is unique up to unique isomorphism. Moreover, suppose that $V \subset Z$ and $U \subset X$ are open subschemes such that $h(V) \subset U$. Let $Z \subset Z'$ be a universal first order thickening of $Z$ over $X$. Let $V' \subset Z'$ be the open subscheme such that $V = Z \cap V'$. Then we claim that $V \subset V'$ is the universal first order thickening of $V$ over $U$. Namely, suppose given any diagram $$ \xymatrix{ V \ar[d]_h & T \ar[l]^a \ar[d] \\ U & T' \ar[l]_b } $$ where $T \subset T'$ is a first order thickening over $U$. By the universal property of $Z'$ we obtain $(a, a') : (T \subset T') \to (Z \subset Z')$. But since we have equality $|T| = |T'|$ of underlying topological spaces we see that $a'(T') \subset V'$. Hence we may think of $(a, a')$ as a morphism of thickenings $(a, a') : (T \subset T') \to (V \subset V')$ over $U$. Uniqueness is clear also. In a completely similar manner one proves that if $h(Z) \subset U$ and $Z \subset Z'$ is a universal first order thickening over $U$, then $Z \subset Z'$ is a universal first order thickening over $X$. \medskip\noindent Before we glue affine pieces let us show that the lemma holds if $Z$ and $X$ are affine. Say $X = \Spec(R)$ and $Z = \Spec(S)$. By Algebra, Lemma \ref{algebra-lemma-universal-thickening} there exists a first order thickening $Z \subset Z'$ over $X$ which has the universal property of the lemma for diagrams $$ \xymatrix{ Z \ar[d]_h & T \ar[l]^a \ar[d] \\ X & T' \ar[l]_b } $$ where $T, T'$ are affine. Given a general diagram we can choose an affine open covering $T' = \bigcup T'_i$ and we obtain morphisms $a'_i : T'_i \to Z'$ over $X$ such that $a'_i|_{T_i} = a|_{T_i}$. By uniqueness we see that $a'_i$ and $a'_j$ agree on any affine open of $T'_i \cap T'_j$. Hence the morphisms $a'_i$ glue to a global morphism $a' : T' \to Z'$ over $X$ as desired. Thus the lemma holds if $X$ and $Z$ are affine. \medskip\noindent Choose an affine open covering $Z = \bigcup Z_i$ such that each $Z_i$ maps into an affine open $U_i$ of $X$. By Lemma \ref{lemma-formally-unramified-on-opens} the morphisms $Z_i \to U_i$ are formally unramified. Hence by the affine case we obtain universal first order thickenings $Z_i \subset Z_i'$ over $U_i$. By the general remarks above $Z_i \subset Z_i'$ is also a universal first order thickening of $Z_i$ over $X$. Let $Z'_{i, j} \subset Z'_i$ be the open subscheme such that $Z_i \cap Z_j = Z'_{i, j} \cap Z_i$. By the general remarks we see that both $Z'_{i, j}$ and $Z'_{j, i}$ are universal first order thickenings of $Z_i \cap Z_j$ over $X$. Thus, by the first of our general remarks, we see that there is a canonical isomorphism $\varphi_{ij} : Z'_{i, j} \to Z'_{j, i}$ inducing the identity on $Z_i \cap Z_j$. We claim that these morphisms satisfy the cocycle condition of Schemes, Section \ref{schemes-section-glueing-schemes}. (Verification omitted. Hint: Use that $Z'_{i, j} \cap Z'_{i, k}$ is the universal first order thickening of $Z_i \cap Z_j \cap Z_k$ which determines it up to unique isomorphism by what was said above.) Hence we can use the results of Schemes, Section \ref{schemes-section-glueing-schemes} to get a first order thickening $Z \subset Z'$ over $X$ which the property that the open subscheme $Z'_i \subset Z'$ with $Z_i = Z'_i \cap Z$ is a universal first order thickening of $Z_i$ over $X$. \medskip\noindent It turns out that this implies formally that $Z'$ is a universal first order thickening of $Z$ over $X$. Namely, we have the universal property for any diagram $$ \xymatrix{ Z \ar[d]_h & T \ar[l]^a \ar[d] \\ X & T' \ar[l]_b } $$ where $a(T)$ is contained in some $Z_i$. Given a general diagram we can choose an open covering $T' = \bigcup T'_i$ such that $a(T_i) \subset Z_i$. We obtain morphisms $a'_i : T'_i \to Z'$ over $X$ such that $a'_i|_{T_i} = a|_{T_i}$. We see that $a'_i$ and $a'_j$ necessarily agree on $T'_i \cap T'_j$ since both $a'_i|_{T'_i \cap T'_j}$ and $a'_j|_{T'_i \cap T'_j}$ are solutions of the problem of mapping into the universal first order thickening $Z'_i \cap Z'_j$ of $Z_i \cap Z_j$ over $X$. Hence the morphisms $a'_i$ glue to a global morphism $a' : T' \to Z'$ over $X$ as desired. This finishes the proof.