Suppose $g,h:Z\to X$ are two morphisms of schemes. Then we say that $f:X\to Y$ is the *coequalizer* of $g$ and $h$ if the following condition holds: any morphism $t:X\to T$ such that $t\circ g=t\circ h$ factors uniquely through $f$. The question is whether it is possible for a coequalizer $f:X\to Y$ to **fail** to be surjective.

**Remark:** $f$ must hit all the *closed* points of $Y$. To see this, suppose $y\in Y$ is a closed point that $f$ misses. Then $f$ factors through the open subscheme $Y\smallsetminus\{y\}$. It is easy to check (using the fact that $Y$ is the coequalizer) that $Y\smallsetminus\{y\}$ satisfies the universal property of the coequalizer. But coequalizers are unique, so we get $Y=Y\smallsetminus\{y\}$.

**Background:** A *categorical quotient* of a scheme $X$ by a group $G$ is the same thing as a coequalizer of the two maps $G\times X\rightrightarrows X$ (given by $(g,x)\mapsto x$ and the action $(g,x)\mapsto g\cdot x$) in the category of schemes. In Geometric Invariant Theory, Mumford defines the notion of a *geometric quotient* of a scheme by a group (Definition 0.6), which is stronger than the notion of a categorical quotient (Proposition 0.1). Part of the definition is that a map $f:X\to Y$ must be surjective in order to be a geometric quotient. In subsequent pages, he suggests strongly (but doesn't explicitly state) that a categorical quotient need not be surjective.