# Concrete proof of "If $U$ and $V$ are open subsets of an $A$-scheme $X$, then $\Delta\cap(U\times_A V)\cong U\cap V$"

This is a snip from Ravi Vakil's notes on algebraic geometry. $$A$$-scheme refers to $$X$$ being a scheme over $$\operatorname{Spec} A$$. Could anybody provide a concrete proof of the proposition mentioned below. $$\Delta$$ refers to the image of the diagonal morphism which is a locally closed subscheme

I will attempt to show that they both equal the product $$U\times_X V$$. This I think should be clear for $$U\cap V$$. We check that $$\Delta\cap (U\times_AV)$$, together ith the natural maps to $$U$$ and $$V$$ have the universal property for $$U\times_X V$$. (We think of $$\Delta\cap (U\times_AV)$$ as a locally closed subscheme of $$X\times_A X$$.) Suppose we have a diagram $$W\to_f V\\ \downarrow_g\quad\quad\downarrow_j\\ U\to_i X$$ Then post composing with the map $$X\to \text{Spec}(A)$$ we get a similar square with $$\text{Spec}(A)$$ in the bottom right corner and therefore a unique map $$W\to U\times_A V$$. Postcomposing this with an open immersion we get a map $$\phi:W\to X\times_A X$$ which lands in $$U\times_A V$$. I will show that this also lands in the image of the diagonal map $$X\to X\times_A X$$. Note that the two maps we get from $$W\to X\times_A X\to X$$ by choosing either of the two projections are equal (they both equal $$i\circ g=j\circ f$$). This map, $$W\to X$$ is such that the composite $$W\to X\to X\times_A X$$ gives us $$\phi$$. . Therefore $$\phi$$ lands in $$\Delta\cap (U\times_A V)$$. So $$\Delta\cap (U\times_A V)$$ hsatisfies the existence part of the universal property of $$U\times_X V$$. Uniqueness should follow from the map $$W\to U\times_A V$$ being unique.