Let $X$ and $Y$ be finite type schemes over $\mathrm{Spec} \mathbb{Z}$ and let $f_\xi : X_\xi \rightarrow Y_\xi$ be a morphism between the generic fibers. Then $f_\xi$ spreads out to a morphism $g_U : X_U \rightarrow Y_U$ over some open $U \subset \mathrm{Spec} \mathbb{Z}$. Furthermore, if $f_\xi$ also spreads out to $h: X_V \rightarrow Y_V$ over $V \subset \mathrm{Spec}\mathbb{Z}$, then $g_W = h_W$ for some non-empty open $W\subset U \cap V$.
My attempt:
I already proved the theorem in the case where both $X$ and $Y$ are affine. Now I'm considering the case only $Y$ affine.
Since $X$ is of finite type we can write $X=\bigcup_{i \in I} \mathrm{Spec\ }A_i$ with $I$ finite and $Y=\mathrm{Spec\ }B$. Now $f_\xi: X_\xi \rightarrow Y_\xi$ and we have for every $j \in I$ the inclusion map $i_j: \mathrm{Spec\ }A_j \rightarrow X$. By the universal property of fiber products there exists unique $i_\xi : \mathrm{Spec\ } A_{i,\xi} \rightarrow X_\xi$. Now we compose every $i_\xi$ with $f_\xi$ to get $f_{i,\xi}=f_\xi \circ i_\xi : \mathrm{Spec\ } A_{i,\xi} \rightarrow Y_\xi$: since now we have reduced to the affine case, every $f_{i,\xi}$ spreads out to a morphism $g_{i,U_i}: \mathrm{Spec\ } A_{i,U_i} \rightarrow Y_{U_i}$ for some $U_i \subset \mathrm{Spec\ }\mathbb{Z}$ open. WLOG we may assume $U_i=U$ for every $I$ since we have finitely many maps. Now I'd like to glue all the $g_{i,U}$'s: for every pair $i,j$ we can cover $\mathrm{Spec}A_{i,U} \cap \mathrm{Spec}A_{j,U}=\bigcup W_k$ by finitely many affine opens and I think it's possible to prove they agree on some open subset $U'_k$ and I can replace $U$ by the intersection of these $U'_k$'s and glue.
Now assume also $Y$ is general, $Y=\bigcup_{j \in J} T_j$, $J$ finite and $T_j$'s open affines. Note that by the universal property of fiber products we have $(\bigcup_{j \in J} T_i)_\xi = \bigcup_{j \in J} (T_i)_\xi$. Now define the maps $f_{j,\xi}: f_{\xi}^{-1}((T_j)_\xi) \rightarrow (T_j)_\xi$. We have to find $Z_j \subset X$ non-empty s.t. $Z_j \cap X_\xi=f_\xi^{-1}((T_j)_\xi)$ so we get $f_{j, \xi}: (Z_j)_\xi \rightarrow (T_j)_\xi$. Now we can spread this out to a morphism $g_j: (Z_j)_{U_j} \rightarrow (T_j)_{U_j}$. Like before we can assume WLOG that $U_j=U$ for all $j$ since there are finitely many maps.Now take the compositions \begin{align*} (Z_j)_U \rightarrow (T_j)_U \rightarrow \bigcup_{j \in J} (T_j)_U=Y_U \end{align*} We call these compositions $h_j$ and we can glue them together to $h: \bigcup_{j \in J} (Z_j)_W \rightarrow Y_U$ for some $W \subset U$.
To conclude we'd like to prove that we can replace $U$ by a smaller non-empty open $W$ to obtain $V_W =X_W$.
