CatDat

An embedding of the category of topological spaces in the category of locally ringed spaces

For much of this development, we will be dealing with the case of $\LRS_k$ where $k$ is a field. We begin by describing $\Top$ as a reflective subcategory of $\LRS_k$.

Proof. In this adjunction, the counit $UK \to \id$ is just the identity. To describe the unit $\id \to KU$, we need to define a morphism $(X, \O_X) \to (X, \underline{k})$ for any locally ringed space $(X, \O_X)$ over $k$. This morphism will be the identity on topological spaces, and the pullback operation $\underline{k} \to \O_X$ will be the unique morphism of sheaves induced by the given structure of $\O_X$ as a sheaf of $k$-algebras. It is now straightforward to check this indeed defines an adjunction; and since the counit is an isomorphism, that implies that $K$ is fully faithful. $\square$

We now show that this reflective subcategory is in fact also a coreflective subcategory. Recall that for $f \in \O_X(U)$ we have its vanishing set $V(f) \coloneqq \{x\in U : f \in \m_{X,x}\} = \{x \in U : f(x) = 0\}$, where $f(x) \in \kappa(x)$ is the image of $f_x \in \O_{X,x}$ in the residue field.

Proof. First, to see that $S$ is a functor, suppose we have a morphism $f : (X, \O_X) \to (Y, \O_Y)$. Then for $x \in X_0$, we have a sequence $k \to \kappa(f(x)) \to \kappa(x)$ where the composition is an isomorphism. Thus, $\kappa(f(x)) \to \kappa(x)$ is a surjective morphism of fields, and therefore an isomorphism. It follows that $k \to \kappa(f(x))$ is also an isomorphism of fields, so $f(x) \in Y_0$. To see that the restriction map $X_0 \to Y_0$ is continuous, suppose $g \in \O_Y(V)$ is such that $f(x) \in V(g)$. Then $x \in V(f^\sharp g)$, and $$X_0 \cap f^{-1}(Y_0 \cap V(g)) = X_0 \cap V(f^\sharp g)$$ where $f^\sharp g \in \O_X(f^{-1}(V))$. In other words, we have shown that the inverse image in $X_0$ of any subbasic neighborhood of $f(x)$ is a neighborhood of $x$.

Now if we apply the functor $S$ to a space of the form $(X, \underline{k})$, then since by definition any section of the constant sheaf $\underline{k}$ is locally constant, we see that we recover exactly $X$ with its original topology. Thus, we can define the unit $\id \to SK$ of the adjunction to be the identity.

As for the counit $KS \to \id$, for any locally ringed space $(X, \O_X)$ over $k$ we need to define a morphism $(X_0, \underline{k}) \to (X, \O_X)$. The map of topological spaces will be the inclusion map $i : X_0 \hookrightarrow X$, which is continuous since in particular for $U$ an open neighborhood of $x \in X_0$ we have $X_0 \cap U = X_0 \cap V(0_U)$, where $0_U \in \O_X(U)$ is the zero element. The pullback map $\O_X \to i_* \underline{k}$ takes $f \in \O_X(U)$ to the function $X_0 \cap U \to k$ where $x \in X_0 \cap U$ maps to the inverse image of $f(x) \in \kappa(x)$ under the isomorphism $k \to \kappa(x)$. An alternative description of this pullback is that $x \in X_0 \cap U$ maps to the unique $a\in k$ such that $x \in V(f-a)$. Since $X_0 \cap V(f-a)$ is a neighborhood of $x$ in $X_0$ by definition, this shows that we get a locally constant function to $k$ as required.

From here, it is straightforward to show that this does in fact define an adjunction. $\square$

Remark. In the special case where $k$ is a finite field, we have $$\textstyle X_0 \cap V(f) = \bigcap_{a \in k^\times} (X_0 \cap D(f-a)),$$ which is already open in the subspace topology. Therefore, in this case, $X_0$ is given exactly the subspace topology.

Proof. The functor $K_R$ is the composition of $K : \Top \to \LRS_k$ and the forgetful functor $\LRS_k \to \LRS_R$. Since $\LRS_k$ is equivalent to the slice category of $\LRS_R$ over the subterminal object $\Spec k$, the forgetful functor is fully faithful; has right adjoint ${-} \times \Spec k$; and preserves all inhabited limits. Therefore, from the previously established results on $K$, the result follows. $\square$

Proof. We already know that $\Top$ does not satisfy any of these properties. In order to conclude that $\LRS_R$ does not satisfy any of them either, we fix a quotient field of $R$ as above and consider the functor $K_R$. In each case, this is an easy application of a contrapositive of a result from here to the functor $K_R$. Namely, (a) follows from Lemma 5; (b) from Lemma 4; (c) from Lemma 7; (d) from the dual of Lemma 2 with the observation that $K_R$ preserves epimorphisms since it has a right adjoint; and (e) from the dual of Lemma 8. $\square$