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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 enter image description here

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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.

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