The Stacks project

Proof. There is a canonical map $|X| \to \mathop{\mathrm{lim}}\nolimits |X_ i|$. Choose $0 \in I$. If $W_0 \subset X_0$ is an open subspace, then we have $f_0^{-1}W_0 = \mathop{\mathrm{lim}}\nolimits _{i \geq 0} f_{i0}^{-1}W_0$, see Lemma 70.4.1. Hence, if we can prove the lemma for inverse systems where $X_0$ is quasi-compact, then the lemma follows in general. Thus we may and do assume $X_0$ is quasi-compact.

Choose an affine scheme $U_0$ and a surjective étale morphism $U_0 \to X_0$. Set $U_ i = X_ i \times _{X_0} U_0$ and $U = X \times _{X_0} U_0$. Set $R_ i = U_ i \times _{X_ i} U_ i$ and $R = U \times _ X U$. Recall that $U = \mathop{\mathrm{lim}}\nolimits U_ i$ and $R = \mathop{\mathrm{lim}}\nolimits R_ i$, see proof of Lemma 70.4.1. Recall that $|X| = |U|/|R|$ and $|X_ i| = |U_ i|/|R_ i|$. By Limits, Lemma 32.4.6 we have $|U| = \mathop{\mathrm{lim}}\nolimits |U_ i|$ and $|R| = \mathop{\mathrm{lim}}\nolimits |R_ i|$.

Surjectivity of $|X| \to \mathop{\mathrm{lim}}\nolimits |X_ i|$. Let $(x_ i) \in \mathop{\mathrm{lim}}\nolimits |X_ i|$. Denote $S_ i \subset |U_ i|$ the inverse image of $x_ i$. This is a finite nonempty set by the definition of decent spaces (Decent Spaces, Definition 68.6.1). Hence $\mathop{\mathrm{lim}}\nolimits S_ i$ is nonempty, see Categories, Lemma 4.21.7. Let $(u_ i) \in \mathop{\mathrm{lim}}\nolimits S_ i \subset \mathop{\mathrm{lim}}\nolimits |U_ i|$. By the above this determines a point $u \in |U|$ which maps to an $x \in |X|$ mapping to the given element $(x_ i)$ of $\mathop{\mathrm{lim}}\nolimits |X_ i|$.

Injectivity of $|X| \to \mathop{\mathrm{lim}}\nolimits |X_ i|$. Suppose that $x, x' \in |X|$ map to the same point of $\mathop{\mathrm{lim}}\nolimits |X_ i|$. Choose lifts $u, u' \in |U|$ and denote $u_ i, u'_ i \in |U_ i|$ the images. For each $i$ let $T_ i \subset |R_ i|$ be the set of points mapping to $(u_ i, u'_ i) \in |U_ i| \times |U_ i|$. This is a finite set by the definition of decent spaces (Decent Spaces, Definition 68.6.1). Moreover $T_ i$ is nonempty as we've assumed that $x$ and $x'$ map to the same point of $X_ i$. Hence $\mathop{\mathrm{lim}}\nolimits T_ i$ is nonempty, see Categories, Lemma 4.21.7. As before let $r \in |R| = \mathop{\mathrm{lim}}\nolimits |R_ i|$ be a point corresponding to an element of $\mathop{\mathrm{lim}}\nolimits T_ i$. Then $r$ maps to $(u, u')$ in $|U| \times |U|$ by construction and we see that $x = x'$ in $|X|$ as desired.

Parenthetical statement: A quasi-separated algebraic space is decent, see Decent Spaces, Section 68.6 (the key observation to this is Properties of Spaces, Lemma 66.6.7). A locally separated algebraic space is decent by Decent Spaces, Lemma 68.15.2. $\square$