## 69.10 Relative approximation

We discuss variants of Proposition 69.8.1 over a base.

Lemma 69.10.1. Let $f : X \to Y$ be a morphism of quasi-compact and quasi-separated algebraic spaces over $\mathbf{Z}$. Then there exists a direct set $I$ and an inverse system $(f_ i : X_ i \to Y_ i)$ of morphisms algebraic spaces over $I$, such that the transition morphisms $X_ i \to X_{i'}$ and $Y_ i \to Y_{i'}$ are affine, such that $X_ i$ and $Y_ i$ are quasi-separated and of finite type over $\mathbf{Z}$, and such that $(X \to Y) = \mathop{\mathrm{lim}}\nolimits (X_ i \to Y_ i)$.

Proof. Write $X = \mathop{\mathrm{lim}}\nolimits _{a \in A} X_ a$ and $Y = \mathop{\mathrm{lim}}\nolimits _{b \in B} Y_ b$ as in Proposition 69.8.1, i.e., with $X_ a$ and $Y_ b$ quasi-separated and of finite type over $\mathbf{Z}$ and with affine transition morphisms.

Fix $b \in B$. By Lemma 69.4.5 applied to $Y_ b$ and $X = \mathop{\mathrm{lim}}\nolimits X_ a$ over $\mathbf{Z}$ we find there exists an $a \in A$ and a morphism $f_{a, b} : X_ a \to Y_ b$ making the diagram

$\xymatrix{ X \ar[d] \ar[r] & Y \ar[d] \\ X_ a \ar[r] & Y_ b }$

commute. Let $I$ be the set of triples $(a, b, f_{a, b})$ we obtain in this manner.

Let $(a, b, f_{a, b})$ and $(a', b', f_{a', b'})$ be in $I$. Let $b'' \leq \min (b, b')$. By Lemma 69.4.5 again, there exists an $a'' \geq \max (a, a')$ such that the compositions $X_{a''} \to X_ a \to Y_ b \to Y_{b''}$ and $X_{a''} \to X_{a'} \to Y_{b'} \to Y_{b''}$ are equal. We endow $I$ with the preorder

$(a, b, f_{a, b}) \geq (a', b', f_{a', b'}) \Leftrightarrow a \geq a',\ b \geq b',\text{ and } g_{b, b'} \circ f_{a, b} = f_{a', b'} \circ h_{a, a'}$

where $h_{a, a'} : X_ a \to X_{a'}$ and $g_{b, b'} : Y_ b \to Y_{b'}$ are the transition morphisms. The remarks above show that $I$ is directed and that the maps $I \to A$, $(a, b, f_{a, b}) \mapsto a$ and $I \to B$, $(a, b, f_{a, b})$ are cofinal. If for $i = (a, b, f_{a, b})$ we set $X_ i = X_ a$, $Y_ i = Y_ b$, and $f_ i = f_{a, b}$, then we get an inverse system of morphisms over $I$ and we have

$\mathop{\mathrm{lim}}\nolimits _{i \in I} X_ i = \mathop{\mathrm{lim}}\nolimits _{a \in A} X_ a = X \quad \text{and}\quad \mathop{\mathrm{lim}}\nolimits _{i \in I} S_ i = \mathop{\mathrm{lim}}\nolimits _{b \in B} Y_ b = Y$

by Categories, Lemma 4.17.4 (recall that limits over $I$ are really limits over the opposite category associated to $I$ and hence cofinal turns into initial). This finishes the proof. $\square$

Lemma 69.10.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume that

1. $X$ is quasi-compact and quasi-separated, and

2. $Y$ is quasi-separated.

Then $X = \mathop{\mathrm{lim}}\nolimits X_ i$ is a limit of a directed inverse system of algebraic spaces $X_ i$ of finite presentation over $Y$ with affine transition morphisms over $Y$.

Proof. Since $|f|(|X|)$ is quasi-compact we may replace $Y$ by a quasi-compact open subspace whose set of points contains $|f|(|X|)$. Hence we may assume $Y$ is quasi-compact as well. By Lemma 69.10.1 we can write $(X \to Y) = \mathop{\mathrm{lim}}\nolimits (X_ i \to Y_ i)$ for some directed inverse system of morphisms of finite type schemes over $\mathbf{Z}$ with affine transition morphisms. Since limits commute with limits (Categories, Lemma 4.14.10) we have $X = \mathop{\mathrm{lim}}\nolimits X_ i \times _{Y_ i} Y$. For $i \geq i'$ the transition morphism $X_ i \times _{Y_ i} Y \to X_{i'} \times _{Y_{i'}} Y$ is affine as the composition

$X_ i \times _{Y_ i} Y \to X_ i \times _{Y_{i'}} Y \to X_{i'} \times _{Y_{i'}} Y$

where the first morphism is a closed immersion (by Morphisms of Spaces, Lemma 66.4.5) and the second is a base change of an affine morphism (Morphisms of Spaces, Lemma 66.20.5) and the composition of affine morphisms is affine (Morphisms of Spaces, Lemma 66.20.4). The morphisms $f_ i$ are of finite presentation (Morphisms of Spaces, Lemmas 66.28.7 and 66.28.9) and hence the base changes $X_ i \times _{f_ i, Y_ i} Y \to Y$ are of finite presentation (Morphisms of Spaces, Lemma 66.28.3). $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).