The Stacks project

110.5 The structure sheaf on the fibre product

Let $X, Y, S, a, b, p, q, f$ be as in the introduction to Derived Categories of Schemes, Section 36.23. Picture:

Then we have a canonical map

which is not an isomorphism in general.

For example, let $S = \mathop{\mathrm{Spec}}(\mathbf{R})$, $X = \mathop{\mathrm{Spec}}(\mathbf{C})$, and $Y = \mathop{\mathrm{Spec}}(\mathbf{C})$. Then $X \times _ S Y = \mathop{\mathrm{Spec}}(\mathbf{C}) \amalg \mathop{\mathrm{Spec}}(\mathbf{C})$ is a discrete space with two points and the sheaves $p^{-1}\mathcal{O}_ X$, $q^{-1}\mathcal{O}_ Y$ and $f^{-1}\mathcal{O}_ S$ are the constant sheaves with values $\mathbf{C}$, $\mathbf{C}$, and $\mathbf{R}$. Hence the source of $can$ is the constant sheaf with value $\mathbf{C} \otimes _\mathbf {R} \mathbf{C}$ on the discrete space with two points. Thus its global sections have dimension $8$ as an $\mathbf{R}$-vector space whereas taking global sections of the target of $can$ we obtain $\mathbf{C} \times \mathbf{C}$ which has dimension $4$ as an $\mathbf{R}$-vector space.

Another example is the following. Let $k$ be an algebraically closed field. Consider $S = \mathop{\mathrm{Spec}}(k)$, $X = \mathbf{A}^1_ k$, and $Y = \mathbf{A}^1_ k$. Then for $U \subset X \times _ S Y = \mathbf{A}^2_ k$ nonempty open the images $p(U) \subset X = \mathbf{A}^1_ k$ and $q(U) \subset \mathbf{A}^1_ k$ are open and the reader can show that

This is not equal to $\mathcal{O}_{X \times _ S Y}(U)$ if $U$ is the complement of an irreducible curve $C$ in $X \times _ S Y = \mathbf{A}^2_ k$ such that both $p|_ C$ and $q|_ C$ are nonconstant.

Returning to the general case, let $z = (x, y, s, \mathfrak p)$ be a point of $X \times _ S Y$ as in Schemes, Lemma 26.17.5. Then on stalks at $z$ the map $can$ gives the map

This is a flat ring homomorphism as the target is a localization of the source (details omitted; hint reduce to the case that $X$, $Y$, and $S$ are affine). Observe that the source is in general not a local ring, and this gives another way to see that $can$ is not an isomorphism in general.

More generally, suppose we have an $\mathcal{O}_ X$-module $\mathcal{F}$ and an $\mathcal{O}_ Y$-module $\mathcal{G}$. Then there is a canonical map

which is rarely an isomorphism.