# Morphisms between affine schemes

Suppose we have two affine schemes $$X=\operatorname{Spec} A$$ and $$Y=\operatorname{Spec} B$$ for commutative rings $$A,B$$. I encountered this statement in my homework that $$\operatorname{Mor}(X,Y)=\operatorname{Hom}(B,A)$$. I understand that a ring homomorphism between $$B$$ and $$A$$ would induce a continuous map between $$X$$ and $$Y$$. But isn't that only for the underlying topological space? Why we don't need to check it is indeed a morphism between the sheaves here?

More generally, if $$X$$ and $$Y$$ are general schemes, is it still enough to specify a continuous map between the underlying topological space to define a morphism between the two schemes?

• First, a ring homomorphism induces a continuous map on topological spaces, not necessarily a homeomorphism. Second, the space $\operatorname{Mor}(X,Y)$ consists of morphisms of schemes, not just continuous maps. Commented Apr 5, 2021 at 15:43
• Thank you for pointing that out. Just edited it. Commented Apr 5, 2021 at 16:20
• My answer here might be helpful. Commented Apr 5, 2021 at 16:28
• Well there is a theorem stating that the map $\alpha$:Hom$($X$,$Y$)$ $\longrightarrow$ Hom$($A(Y)$,$A(X)$)$ is a homeomorphism if $X$ is any variety and $Y$ is affine. (I have no clue if this will help. I just wanted to mention something similar. Also recall that $A$ is actually a functor and that the inverse functor of $A$ is $Spec$). Commented Apr 5, 2021 at 16:55

A map $$B \longrightarrow A$$ does not generally describe a homeomorphism $$\operatorname{Spec} A \longrightarrow \operatorname{Spec} B$$, but it does describe a continuous map. For instance, the map $$\mathbb Z \longrightarrow \mathbb Z/2$$ corresponds to the inclusion $$\{(2)\} \subseteq \operatorname{Spec} \mathbb Z$$ - certainly not a homeomorphism.
Also, you're right that a map of schemes requires a map of the structure sheaves. This is not derivable from the map on the underlying space. Instead it comes from the map of rings. Indeed, the idea is that for a basic open set $$D(f) \subseteq \operatorname{Spec} B$$, its pullback under the map induced by $$\phi: B \longrightarrow A$$ is going to be $$D(\phi(f))$$. Recall that $$\Gamma(D(f), \operatorname{Spec} B) = B_f$$. We're looking for a map $$\Gamma(D(f), \operatorname{Spec} B) \longrightarrow \Gamma(D(\phi(f)), \operatorname{Spec} A)$$ to define our map of structure sheaves. We'll then take this to be the map $$B_f \longrightarrow A_{\phi(f)}$$, which we get from the universal property of localization applied to $$\phi$$. Now, I've only defined the map on a basis of open sets, but general properties of sheaves imply that this is enough. Also, you need to check that the induced maps on stalks are all local, but I won't show this. Here is a reference to the Stacks project for this fact.
Now I hope this construction explains why a ring homomorphism induces a reverse map of affine schemes, structure sheaf and all. But I did say that you cannot derive this map purely from the map on the underlying topological spaces, and I'd like to explain that further. Indeed, in defining a map of affine schemes it would suffice to just specify a continuous map if, for instance, the forgetful functor $$F: \mathbf{AffineSchemes} \longrightarrow \mathbf{Top}$$ was fully faithful. In fact, it is neither full mor faithful.
For fullness, consider maps $$\operatorname{Spec} \mathbb Z \longrightarrow \operatorname{Spec} \mathbb Z$$. There is only one ring homomorphism $$\mathbb Z \longrightarrow \mathbb Z$$, so the only map between the affine schemes $$\operatorname{Spec} \mathbb Z \longrightarrow \operatorname{Spec} \mathbb Z$$ is the identity. However, there are many continuous maps between the underlying spaces $$F(\operatorname{Spec} \mathbb Z) \longrightarrow F(\operatorname{Spec} \mathbb Z)$$. For instance, there are infinitely many constant maps, all of which are continuous. Hence, the forgetful functor $$F$$ cannot be full. In other words, not every map of the underlying topological space can arise from a map of affine schemes.
Now for faithfulness, consider $$\mathbb Q(i)$$. We have two automorphisms of this field, the identity and complex conjugation. They therefore define two automorphisms of the affine scheme $$\operatorname{Spec} \mathbb Q(i)$$. However, $$\mathbb Q(i)$$ is a field so the topological space of the affine scheme, $$F(\operatorname{Spec} \mathbb Q(i))$$, is a single point. There is only one continuous map from a point to itself, so $$F$$ cannot be faithful. That is, it is impossible in general to take a continuous map between affine schemes and derive a corresponding map on the structure sheaves as the same continuous map can come from many scheme maps.
These examples also show that the forgetful functor $$\mathbf{Schemes} \longrightarrow \mathbf{Top}$$ is neither full nor faithful, so you cannot define a map of schemes purely from a continuous map. I'd also like to point out that unlike affine schemes, a map between general schemes is not determined by a (reversed direction) ring homomorphism on global sections. Indeed, consider projective space over an algebraically closed field $$k$$. The global sections of $$\mathbb P_k^n$$ is $$k$$ for all $$n$$. There are many maps between projective spaces that are constant on global sections. For instance, linear automorphisms of $$\mathbb P_k^n$$ are defined by matrices in $$PGL_n(k)$$, and in general this group is nontrivial.