# 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.

## 1 Answer

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 $$f\neq g$$.