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](http://books.google.com/books?id=dFlv3zn_2-gC), 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.



Can a coequalizer of schemes fail to be surjective?
Let $k$ be a field. Take $Y=\mathrm{Spec}\,k[[t]]$, and take for $X$ the disjoint sum of the closed subschemes $X\_n:=\mathrm{Spec}\,k[[t]]/(t^n)$ ($n>0$). Put $Z=X\times\_Y X$ with the two obvious maps to $X$. A coequalizer is just a direct limit of the system $X\_1\hookrightarrow\dots X\_n\hookrightarrow X\_{n+1}\hookrightarrow\dots$ in the category of schemes (look at the definition!). Clearly, $Y$ is a direct limit in the category of *affine* schemes, hence also in the category of schemes since the $X\_n$'s are one-point schemes and every compatible system of morphisms $(X\_n\to T)\_{n>0}$ must factor through an affine open subscheme of $T$. 


So, $X\to Y$ is a coequalizer of $pr\_1, pr\_2 :Z\to X$, but its set-theoretic image is the closed point.


