The Stacks project

Proof. Let $x_1, x_2, \ldots $ be a countable set of generators for $M$. For each $x_ n$ choose $i \in I$ such that $x_ n$ is in the image of the canonical map $f_ i: M_ i \to M$; let $I'_{0} \subset I$ be the set of all these $i$. Now since $M$ is Mittag-Leffler, for each $i \in I'_{0}$ we can choose $j \in I$ such that $j \geq i$ and $f_{ij}: M_ i \to M_ j$ factors through $f_{ik}: M_ i \to M_ k$ for all $k \geq i$ (condition (3) of Proposition 10.88.6); let $I'_1$ be the union of $I'_0$ with all of these $j$. Since $I'_1$ is a countable, we can enlarge it to a countable directed set $I'_{2} \subset I$. Now we can apply the same procedure to $I'_{2}$ as we did to $I'_{0}$ to get a new countable set $I'_{3} \subset I$. Then we enlarge $I'_{3}$ to a countable directed set $I'_{4}$. Continuing in this way—adding in a $j$ as in Proposition 10.88.6 (3) for each $ i \in I'_{\ell }$ if $\ell $ is odd and enlarging $I'_{\ell }$ to a directed set if $\ell $ is even—we get a sequence of subsets $I'_{\ell } \subset I$ for $\ell \geq 0$. The union $I' = \bigcup I'_{\ell }$ satisfies:

1. $I'$ is countable and directed;

2. each $x_ n$ is in the image of $f_ i: M_ i \to M$ for some $i \in I'$;

3. if $i \in I'$, then there is $j \in I'$ such that $j \geq i$ and $f_{ij}: M_ i \to M_ j$ factors through $f_{ik}: M_ i \to M_ k$ for all $k \in I$ with $k \geq i$. In particular $\mathop{\mathrm{Ker}}(f_{ik}) \subset \mathop{\mathrm{Ker}}(f_{ij})$ for $k \geq i$.

We claim that the canonical map $\mathop{\mathrm{colim}}\nolimits _{i \in I'} M_ i \to \mathop{\mathrm{colim}}\nolimits _{i \in I} M_ i = M$ is an isomorphism. By (2) it is surjective. For injectivity, suppose $x \in \mathop{\mathrm{colim}}\nolimits _{i \in I'} M_ i$ maps to $0$ in $\mathop{\mathrm{colim}}\nolimits _{i \in I} M_ i$. Representing $x$ by an element $\tilde{x} \in M_ i$ for some $i \in I'$, this means that $f_{ik}(\tilde{x}) = 0$ for some $k \in I, k \geq i$. But then by (3) there is $j \in I', j \geq i,$ such that $f_{ij}(\tilde{x}) = 0$. Hence $x = 0$ in $\mathop{\mathrm{colim}}\nolimits _{i \in I'} M_ i$. $\square$