Lemma 69.23.4. In Situation 69.23.1. Let $f : X \to Y$ be a morphism of algebraic spaces quasi-separated and of finite type over $B$. Let

$\vcenter { \xymatrix{ X \ar[r] \ar[d] & W \ar[d] \\ B \ar[r] & B_{i_1} } } \quad \text{and}\quad \vcenter { \xymatrix{ Y \ar[r] \ar[d] & V \ar[d] \\ B \ar[r] & B_{i_2} } }$

be diagrams as in (69.23.2.1). Let $X = \mathop{\mathrm{lim}}\nolimits _{i \geq i_1} X_ i$ and $Y = \mathop{\mathrm{lim}}\nolimits _{i \geq i_2} Y_ i$ be the corresponding limit descriptions as in Lemma 69.23.3. Then there exists an $i_0 \geq \max (i_1, i_2)$ and a morphism

$(f_ i)_{i \geq i_0} : (X_ i)_{i \geq i_0} \to (Y_ i)_{i \geq i_0}$

of inverse systems over $(B_ i)_{i \geq i_0}$ such that such that $f = \mathop{\mathrm{lim}}\nolimits _{i \geq i_0} f_ i$. If $(g_ i)_{i \geq i_0} : (X_ i)_{i \geq i_0} \to (Y_ i)_{i \geq i_0}$ is a second morphism of inverse systems over $(B_ i)_{i \geq i_0}$ such that such that $f = \mathop{\mathrm{lim}}\nolimits _{i \geq i_0} g_ i$ then $f_ i = g_ i$ for all $i \gg i_0$.

Proof. Since $V \to B_{i_2}$ is of finite presentation and $X = \mathop{\mathrm{lim}}\nolimits _{i \geq i_1} X_ i$ we can appeal to Proposition 69.3.10 as improved by Lemma 69.4.5 to find an $i_0 \geq \max (i_1, i_2)$ and a morphism $h : X_{i_0} \to V$ over $B_{i_2}$ such that $X \to X_{i_0} \to V$ is equal to $X \to Y \to V$. For $i \geq i_0$ we get a commutative solid diagram

$\xymatrix{ X \ar[d] \ar[r] & X_ i \ar[r] \ar@{..>}[d] \ar@/_2pc/[dd] |!{[d];[ld]}\hole & X_{i_0} \ar[d]^ h \\ Y \ar[r] \ar[d] & Y_ i \ar[r] \ar[d] & V \ar[d] \\ B \ar[r] & B_ i \ar[r] & B_{i_0} }$

Since $X \to X_ i$ has scheme theoretically dense image and since $Y_ i$ is the scheme theoretic image of $Y \to B_ i \times _{B_{i_2}} V$ we find that the morphism $X_ i \to B_ i \times _{B_{i_2}} V$ induced by the diagram factors through $Y_ i$ (Morphisms of Spaces, Lemma 66.16.6). This proves existence.

Uniqueness. Let $E_ i \to X_ i$ be the equalizer of $f_ i$ and $g_ i$ for $i \geq i_0$. We have $E_ i = Y_ i \times _{\Delta , Y_ i \times _{B_ i} Y_ i, (f_ i, g_ i)} X_ i$. Hence $E_ i \to X_ i$ is a monomorphism of finite presentation as a base change of the diagonal of $Y_ i$ over $B_ i$, see Morphisms of Spaces, Lemmas 66.4.1 and 66.28.10. Since $X_ i$ is a closed subspace of $B_ i \times _{B_{i_0}} X_{i_0}$ and similarly for $Y_ i$ we see that

$E_ i = X_ i \times _{(B_ i \times _{B_{i_0}} X_{i_0})} (B_ i \times _{B_{i_0}} E_{i_0}) = X_ i \times _{X_{i_0}} E_{i_0}$

Similarly, we have $X = X \times _{X_{i_0}} E_{i_0}$. Hence we conclude that $E_ i = X_ i$ for $i$ large enough by Lemma 69.6.10. $\square$

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).