If the equalizer of two scheme morphisms is surjective, does that mean the two morphisms are the same? Let $Z$ be a scheme, and $f , g : X \rightarrow Y$ be two morphisms of schemes over $Z$. Suppose the continuous function of $i : eq(f,g) \rightarrow X$ is surjective. Does this imply that $f  = g$?
My guess is no, because the functor sending a scheme to its underlying set of points is very badly behaved, and scheme morphisms are not determined completely by their underlying continuous function.
My idea for a counter-example is that: if the underlying continuous function $eq(f, g)$ had the same topological space as $X$ but different structure sheaves, then the map $eq(f, g) \rightarrow X$ is not the identity but is surjective. 
One way to get $eq(f, g)$ and $X$ having the same underlying space is to have $X = Spec(A)$ be affine, and $eq(f, g) \rightarrow X$ being the map $A \rightarrow A / nil(A)$. However, that would mean I have to construct it as a coequalizer, and I don't know how to do that. Furthermore, I'm not sure if the equalizer in the category of affine schemes is the same as the equalizer in the category of all schemes. 
 A: First, note that the functor $\operatorname{Spec}:\mathtt{CRing}^{op}\to\mathtt{Sch}$ has a left adjoint, namely the functor that sends a scheme to its ring of global functions (proof: a morphism $X\to\operatorname{Spec} A$ is equivalent to a compatible family of homomorphisms $A\to B$ for each affine open $\operatorname{Spec} B$ in $X$, which by gluing is equivalent to a homomorphism $A\to\mathcal{O}_X(X)$).  Thus $\operatorname{Spec}$ preserves limits, and in particular we can compute equalizers of affine schemes by just taking coequalizers of rings.
Now for an example. Let $k$ be a field, let $A=k[x]/(x^2)$, and let $f^*,g^*:A\to A$ be the identity map and the map that sends $x$ to $0$, respectively.  Letting $X=\operatorname{Spec} A$, these induce morphisms $f,g:X\to X$.  The coequalizer of $f^*$ and $g^*$ is just the obvious quotient map $k[x]/(x^2)\to k$ and so the equalizer of $f$ and $g$ is the corresponding map $i:\operatorname{Spec} k\to X$. This map $i$ is surjective, but $f\neq g$.
