Lemma 87.29.3. Let $S$ be a scheme. Let $X$ be an affine formal algebraic space over $S$. Assume $X$ is McQuillan and let $A$ be the weakly admissible topological ring associated to $X$. Then there is an anti-equivalence of categories between

1. the category $\mathcal{C}$ introduced above, and

2. the category of maps $Y \to X$ of finite type of affine formal algebraic spaces.

Proof. Let $(I_\lambda )$ be a fundamental system of weakly admissible ideals of definition in $A$. Consider $Y$ as in (2). Then $Y \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda )$ is affine (Definition 87.24.1 and Lemma 87.19.7). Say $Y \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda ) = \mathop{\mathrm{Spec}}(B_\lambda )$. The ring map $A/I_\lambda \to B_\lambda$ is of finite type because $\mathop{\mathrm{Spec}}(B_\lambda ) \to \mathop{\mathrm{Spec}}(A/I_\lambda )$ is of finite type (by Definition 87.24.1). Then $(B_\lambda )$ is an object of $\mathcal{C}$.

Conversely, given an object $(B_\lambda )$ of $\mathcal{C}$ we can set $Y = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(B_\lambda )$. This is an affine formal algebraic space. We claim that

$Y \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda ) = \left(\mathop{\mathrm{colim}}\nolimits _\mu \mathop{\mathrm{Spec}}(B_\mu )\right) \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda ) = \mathop{\mathrm{Spec}}(B_\lambda )$

To show this it suffices we get the same values if we evaluate on a quasi-compact scheme $U$. A morphism $U \to \left(\mathop{\mathrm{colim}}\nolimits _\mu \mathop{\mathrm{Spec}}(B_\mu )\right) \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda )$ comes from a morphism $U \to \mathop{\mathrm{Spec}}(B_\mu ) \times _{\mathop{\mathrm{Spec}}(A/I_\mu )} \mathop{\mathrm{Spec}}(A/I_\lambda )$ for some $\mu \geq \lambda$ (use Lemma 87.9.4 two times). Since $\mathop{\mathrm{Spec}}(B_\mu ) \times _{\mathop{\mathrm{Spec}}(A/I_\mu )} \mathop{\mathrm{Spec}}(A/I_\lambda ) = \mathop{\mathrm{Spec}}(B_\lambda )$ by our second assumption on objects of $\mathcal{C}$ this proves what we want. Using this we can show the morphism $Y \to X$ is of finite type. Namely, we note that for any morphism $U \to X$ with $U$ a quasi-compact scheme, we get a factorization $U \to \mathop{\mathrm{Spec}}(A/I_\lambda ) \to X$ for some $\lambda$ (see lemma cited above). Hence

$Y \times _ X U = Y \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda )) \times _{\mathop{\mathrm{Spec}}(A/I_\lambda )} U = \mathop{\mathrm{Spec}}(B_\lambda ) \times _{\mathop{\mathrm{Spec}}(A/I_\lambda )} U$

is a scheme of finite type over $U$ as desired. Thus the construction $(B_\lambda ) \mapsto \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(B_\lambda )$ does give a functor from category (1) to category (2).

To finish the proof we show that the above constructions define quasi-inverse functors between the categories (1) and (2). In one direction you have to show that

$\left(\mathop{\mathrm{colim}}\nolimits _\mu \mathop{\mathrm{Spec}}(B_\mu )\right) \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda ) = \mathop{\mathrm{Spec}}(B_\lambda )$

for any object $(B_\lambda )$ in the category $\mathcal{C}$. This we proved above. For the other direction you have to show that

$Y = \mathop{\mathrm{colim}}\nolimits (Y \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda ))$

given $Y$ in the category (2). Again this is true by evaluating on quasi-compact test objects and because $X = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A/I_\lambda )$. $\square$

Comment #1970 by Brian Conrad on

In the text preceding this lemma, "$I_{\lambda}$" should be "$(I_{\lambda})$", "systems $(B_{\lambda})$" should be "inverse systems $(B_{\lambda})$", and "systems of homomorphisms" should be "compatible systems of homomorphisms".

On the first line of the proof, "$I_{\lambda}$" should be "$(I_{\lambda})$", and just after the first sentence insert "Consider $Y$ as in (2)".

A bit more seriously, since the details are omitted at the end I wonder how you justify that the two constructions are really inverse to each other when $Y$ isn't McQuillan. That is, if there is a counterexample to part 3 of the subsequent Remark (Tag 0AJK) then wouldn't we run into a problem when starting with (1), making the construction in (2), and then coming back to a construction as in (1) (i.e., seems we wouldn't get what we start with).

Comment #1979 by on

OK, I wrote out the details to make sure the construction works. See here or wait till I update the site later today.

Comment #1980 by jojo on

At the end there is a type quasi-comnpact should be quasi-compact.

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