The Stacks project

Proof. Please skip this proof. It is more important to learn how to work with the fibre product which is explained in the next section.

By Lemma 26.6.4 the scheme $\mathop{\mathrm{Spec}}(\mathbf{Z})$ is a final object in the category of locally ringed spaces. Thus it suffices to prove that fibred products exist.

Let $f : X \to S$ and $g : Y \to S$ be morphisms of schemes. We have to show that the functor

is representable. We claim that Lemma 26.15.4 applies to the functor $F$. If we prove this then the lemma is proved.

First we show that $F$ satisfies the sheaf property in the Zariski topology. Namely, suppose that $T$ is a scheme, $T = \bigcup _{i \in I} U_ i$ is an open covering, and $\xi _ i \in F(U_ i)$ such that $\xi _ i|_{U_ i \cap U_ j} = \xi _ j|_{U_ i \cap U_ j}$ for all pairs $i, j$. By definition $\xi _ i$ corresponds to a pair $(a_ i, b_ i)$ where $a_ i : U_ i \to X$ and $b_ i : U_ i \to Y$ are morphisms of schemes such that $f \circ a_ i = g \circ b_ i$. The glueing condition says that $a_ i|_{U_ i \cap U_ j} = a_ j|_{U_ i \cap U_ j}$ and $b_ i|_{U_ i \cap U_ j} = b_ j|_{U_ i \cap U_ j}$. Thus by glueing the morphisms $a_ i$ we obtain a morphism of locally ringed spaces (i.e., a morphism of schemes) $a : T \to X$ and similarly $b : T \to Y$ (see for example the mapping property of Lemma 26.14.1). Moreover, on the members of an open covering the compositions $f \circ a$ and $g \circ b$ agree. Therefore $f \circ a = g \circ b$ and the pair $(a, b)$ defines an element of $F(T)$ which restricts to the pairs $(a_ i, b_ i)$ on each $U_ i$. The sheaf condition is verified.

Next, we construct the family of subfunctors. Choose an open covering by open affines $S = \bigcup \nolimits _{i \in I} U_ i$. For every $i \in I$ choose open coverings by open affines $f^{-1}(U_ i) = \bigcup \nolimits _{j \in J_ i} V_ j$ and $g^{-1}(U_ i) = \bigcup \nolimits _{k \in K_ i} W_ k$. Note that $X = \bigcup _{i \in I} \bigcup _{j \in J_ i} V_ j$ is an open covering and similarly for $Y$. For any $i \in I$ and each pair $(j, k) \in J_ i \times K_ i$ we have a commutative diagram

where all the skew arrows are open immersions. For such a triple we get a functor

There is an obvious transformation of functors $F_{i, j, k} \to F$ (coming from the huge commutative diagram above) which is injective, so we may think of $F_{i, j, k}$ as a subfunctor of $F$.

We check condition (2)(a) of Lemma 26.15.4. This follows directly from Lemma 26.6.7. (Note that we use here that the fibre products in the category of affine schemes are also fibre products in the whole category of locally ringed spaces.)

We check condition (2)(b) of Lemma 26.15.4. Let $T$ be a scheme and let $\xi \in F(T)$. In other words, $\xi = (a, b)$ where $a : T \to X$ and $b : T \to Y$ are morphisms of schemes such that $f \circ a = g \circ b$. Set $V_{i, j, k} = a^{-1}(V_ j) \cap b^{-1}(W_ k)$. For any further morphism $h : T' \to T$ we have $h^*\xi = (a \circ h, b \circ h)$. Hence we see that $h^*\xi \in F_{i, j, k}(T')$ if and only if $a(h(T')) \subset V_ j$ and $b(h(T')) \subset W_ k$. In other words, if and only if $h(T') \subset V_{i, j, k}$. This proves condition (2)(b).

We check condition (2)(c) of Lemma 26.15.4. Let $T$ be a scheme and let $\xi = (a, b) \in F(T)$ as above. Set $V_{i, j, k} = a^{-1}(V_ j) \cap b^{-1}(W_ k)$ as above. Condition (2)(c) just means that $T = \bigcup V_{i, j, k}$ which is evident. Thus the lemma is proved and fibre products exist. $\square$