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.