# Does pullback of schemes by monomorphism produce topological pullback?

Suppose I have scheme maps $X\to Z,Y\to Z$. It is not in general true that the fibre product $X\times_{Z}Y$ has the same underlying topological space (or even the same underlying set) as the fiber product of topological spaces, e.g. taking both maps to be $Spec(\mathbb{C})\to Spec(\mathbb{R})$.

However, I believe it is not hard to show (basically one computes the fibre product affine-locally) that the underlying space of $X\times_{Z}Y$ IS homeomorphic to the topological pullback if $X\to Z$ is any of the following:

• an open embedding
• a closed embedding
• induced by localization (i.e. $Spec(S^{-1}A)\to Spec(A)$.)

In particular, if one composes these things in the correct order, I believe this gives a one-line proof that the scheme-theoretic fibre over a point agrees with the topological fibre.

These also happen to be the canonical examples of monomorphisms (along with compositions of these) in the category of schemes that I know of. So my question is:

If $X\to Z$ is a monomorphism, is the underlying space of $X\times_{Z}Y$ homeomorphic to the pullback in the category of topological spaces?

Yes.

By the the explicit construction of the fiber product of locally ringed spaces (in particular, of schemes), it follows that the continuous map $|X \times_Z Y|\to |X| \times_{|Z|} |Y|$ is surjective and that the fiber over some point $(x,y,z)$ in $|X| \times_{|Z|} |Y|$ is $\mathrm{Spec}(\kappa(x) \otimes_{\kappa(z)} \kappa(y))$. A basis of the topology on $|X \times_Z Y|$ is given by the open subsets $$\Omega(U,V,T,f) = \{(x,y,z,\mathfrak{p}) : x \in U, y \in V, z \in T, f(x,y,z) \notin \mathfrak{p}\}$$ where $U \subseteq X$, $V \subseteq Y$, $T \subseteq Z$ are open subsets such that $U$ and $V$ map into $T$, $f \in \mathcal{O}_X(U) \otimes_{\mathcal{O}_Z(T)} \mathcal{O}_Y(V)$, and $f(x,y,z)$ denotes the image of $f$ in $\kappa(x) \otimes_{\kappa(z)} \kappa(y)$.

A reference for the theory of monomorphisms of schemes resp. epimorphisms of commutative rings is

Séminaire Samuel. Algèbre commutative, 2, 1967-1968, Les épimorphismes d'anneaux, available at nundam.

By Prop. 1.5 in Lazard's Exp. No. 4 a monomorphism $X \to Z$ is injective on the underlying sets and induces isomorphisms on residue fields. Therefore, $\mathrm{Spec}(\kappa(x) \otimes_{\kappa(z)} \kappa(y)) = \mathrm{Spec}(\kappa(y))$ is a single point. It follows that the continuous map $|X \times_Z Y|\to |X| \times_{|Z|} |Y|$ is bijective. The image of a basic-open subset $\Omega(U,V,T,f)$ is $$\{(x,y,z) \in |U| \times_{|T|} |V| : f(x,y,z) \neq 0\}.$$ It is open: If $f(x,y,z) \neq 0$, then $f_{x,y,z} \in \mathcal{O}_{X,x} \otimes_{\mathcal{O}_{Z,z}} \mathcal{O}_{Y,y}$ is invertible. Since directed colimits commute with tensor products, we infer that there are open neighborhoods $U',V',T'$ of $x,y,z$ inside $U,V,T$ such that a) the inverse $f^{-1}$ of $f$ is defined in $\mathcal{O}_X(U') \otimes_{\mathcal{O}_Z(T')} \mathcal{O}_Y(V')$, and b) the equation $f f^{-1} = 1$ already holds in this tensor product. It follows that $U' \times_{V'} T'$ is contained in the set, and this is open with respect to the fiber product topology.

• Wonderful! This is a surprisingly clean description of the fibre product that I wasn't aware of. The reference for monomorphisms is helpful as well. Thanks!
– LCL
Commented Jul 15, 2014 at 13:18