Restricting fibre products to open subschemes

Given: The objects, hypotheses and conventions in the statement above.

Let $f:X\to S$ and $g:Y\to S$ be morphisms of schemes. A fibre product is a scheme $P$, with projections $p:P\to X$ and $q:P\to Y$, such that $fp=gq$ and, for every scheme $T$ and morphisms $a:T\to X$, $b:T\to Y$ with $fa=gb$, there is exactly one $h:T\to P$ satisfying $ph=a$ and $qh=b$. Thus, naturally in every test scheme $T$, $\mathrm{Hom}(T,P)\cong \mathrm{Hom}(T,X){\times }_{\mathrm{Hom}(T,S)}\mathrm{Hom}(T,Y)\mathrm{.}$ Write $P=X{\times }_{S}Y$. The commutative square with edges $p,q,f,g$ is Cartesian when it has this universal property. Morphisms here are morphisms of locally ringed spaces, as in def-morphism-of-schemes. No existence assertion is part of the definition. (Fibre product of schemes)

A morphism $j:U\to X$ is an open immersion if it identifies $U$ isomorphically with an open subscheme of $X$. (Open immersions of schemes)

An open immersion is a monomorphism of schemes, and a composite of open immersions is an open immersion. (Open immersions are monomorphisms)

If $(P,p,q)$ and $(P^{\mathrm{\prime }},p^{\mathrm{\prime }},q^{\mathrm{\prime }})$ are fibre products of the same pair $X\to S\leftarrow Y$, there is a unique isomorphism $u:P\to P^{\mathrm{\prime }}$ with $p^{\mathrm{\prime }}u=p$ and $q^{\mathrm{\prime }}u=q$. (Uniqueness of the fibre product)