Monomorphisms and diagonals

By [F2] a morphism $h:T\to X{\times }_{Y}X$ corresponds to the pair $({\mathrm{pr}}_{1}h,{\mathrm{pr}}_{2}h)$ of morphisms $T\to X$ with $j{\mathrm{pr}}_{1}h=j{\mathrm{pr}}_{2}h$, and $h$ is determined by that pair; the diagonal ${\mathrm{\Delta }}_{X\mathrm{/}Y}$ corresponds to the pair $({\mathrm{id}}_X,{\mathrm{id}}_X)$.

By [F3] the morphism $j$ is a monomorphism exactly when for every scheme $T$ and all $a,b:T\to X$ with $ja=jb$ one has $a=b$.

Assume that $j$ is a monomorphism and let $h:T\to X{\times }_{Y}X$. The pair $(a,b)=({\mathrm{pr}}_{1}h,{\mathrm{pr}}_{2}h)$ satisfies $ja=jb$, so $a=b$ by [F3]; hence ${\mathrm{pr}}_{1}h={\mathrm{pr}}_{2}h$ for every $h$, and taking $T=X{\times }_{Y}X$ and $h=\mathrm{id}$ gives ${\mathrm{pr}}_1={\mathrm{pr}}_2$. Consequently ${\mathrm{pr}}_1{\mathrm{\Delta }}_{X\mathrm{/}Y}={\mathrm{id}}_X$ and also ${\mathrm{pr}}_2{\mathrm{\Delta }}_{X\mathrm{/}Y}={\mathrm{id}}_X$, so ${\mathrm{\Delta }}_{X\mathrm{/}Y}{\mathrm{pr}}_1$ and ${\mathrm{id}}_{X{\times }_{Y}X}$ are two morphisms with the same two projections, equal by the uniqueness in [F2]; hence ${\mathrm{\Delta }}_{X\mathrm{/}Y}$ is an isomorphism with inverse ${\mathrm{pr}}_1$.

Assume that ${\mathrm{\Delta }}_{X\mathrm{/}Y}$ is an isomorphism and let $a,b:T\to X$ satisfy $ja=jb$. By [F2] there is $h:T\to X{\times }_{Y}X$ with ${\mathrm{pr}}_{1}h=a$ and ${\mathrm{pr}}_{2}h=b$. Then $h={\mathrm{\Delta }}_{X\mathrm{/}Y}\circ ({\mathrm{pr}}_1\circ h)$, because ${\mathrm{pr}}_1{\mathrm{\Delta }}_{X\mathrm{/}Y}={\mathrm{id}}_X$ by [F1] and the assumed invertibility of ${\mathrm{\Delta }}_{X\mathrm{/}Y}$ imply ${\mathrm{pr}}_1={\mathrm{\Delta }}_{X\mathrm{/}Y}^{-1}$, so $b={\mathrm{pr}}_{2}h={\mathrm{pr}}_2{\mathrm{\Delta }}_{X\mathrm{/}Y}{\mathrm{pr}}_{1}h=a$. Hence $j$ is a monomorphism.

If $j$ is a monomorphism, step 2.1 exhibits ${\mathrm{\Delta }}_{X\mathrm{/}Y}$ as an isomorphism, and an isomorphism is a closed immersion: its underlying map is a homeomorphism onto its full image and the structure map is an isomorphism of sheaves, hence surjective. By [F4] the monomorphism $j$ is therefore separated.

Steps 2.1 and 2.2 prove the equivalence and step 3.1 the final assertion; note that separatedness here is that of the morphism $j$ with diagonal over $Y$, and nothing is asserted about separatedness of $X$ over other bases.