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We first justify surjectivity is preserved by base change for morphisms of schemes. Given $y^{\prime }$ over $y$ and a point $x$ over $y$, the fiber after the residue-field extension contains a point becauseIndeed, tensoring a nonzero vector space over a field with a field extension remains nonzero; a nonzero commutative ring has a prime ideal, giving the required point. Thus every base change of $f$ is surjective.

Now make an arbitrary base change $Z^{\prime }\to Z$. For a closed subset $C\subseteq Y^{\prime }=Y{\times }_Z{Z}^{\prime }$, its inverse image under $f^{\prime }:X^{\prime }\to Y^{\prime }$ is closed. The proper morphism $(gf{)}^{\prime }:X^{\prime }\to Z^{\prime }$ therefore has closed image on that inverse image. Surjectivity of $f^{\prime }$ givesso $g^{\prime }(C)$ is closed. This proves that $g$ is a universally closed morphism. Together with the given separatedness and finite type, it proves

Let $f:X\to Y$ be a surjective morphism of schemes and let $Y^{\prime }\to Y$ be any morphism of schemes. Take any point $y^{\prime }\in Y^{\prime }$ and let $y$ be its image in $Y$. Surjectivity supplies $x\in X$ with $f(x)=y$. By the point-lifting property of a scheme fibre product proved in part (c), there is a point of $X{\times }_Y{Y}^{\prime }$ projecting to both $x$ and $y^{\prime }$. Thus every point of $Y^{\prime }$ is hit by the projection.

This proves that surjectivity is preserved by base change. No restriction is imposed on the base change of a morphism of schemes. No flatness or finite-type assumption is required.