Proposition 10.88.6. Let $M$ be an $R$-module. Let $(M_ i, f_{ij})$ be a directed system of finitely presented $R$-modules, indexed by $I$, such that $M = \mathop{\mathrm{colim}}\nolimits M_ i$. Let $f_ i: M_ i \to M$ be the canonical map. The following are equivalent:

1. For every finitely presented $R$-module $P$ and module map $f: P \to M$, there exists a finitely presented $R$-module $Q$ and a module map $g: P \to Q$ such that $g$ and $f$ dominate each other, i.e., $\mathop{\mathrm{Ker}}(f \otimes _ R \text{id}_ N) = \mathop{\mathrm{Ker}}(g \otimes _ R \text{id}_ N)$ for every $R$-module $N$.

2. For each $i \in I$, there exists $j \geq i$ such that $f_{ij}: M_ i \to M_ j$ dominates $f_ i: M_ i \to M$.

3. For each $i \in I$, there exists $j \geq i$ such that $f_{ij}: M_ i \to M_ j$ factors through $f_{ik}: M_ i \to M_ k$ for all $k \geq i$.

4. For every $R$-module $N$, the inverse system $(\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, N), \mathop{\mathrm{Hom}}\nolimits _ R(f_{ij}, N))$ is Mittag-Leffler.

5. For $N = \prod _{s \in I} M_ s$, the inverse system $(\mathop{\mathrm{Hom}}\nolimits _ R(M_ i, N), \mathop{\mathrm{Hom}}\nolimits _ R(f_{ij}, N))$ is Mittag-Leffler.

Proof. First we prove the equivalence of (1) and (2). Suppose (1) holds and let $i \in I$. Corresponding to the map $f_ i: M_ i \to M$, we can choose $g: M_ i \to Q$ as in (1). Since $M_ i$ and $Q$ are of finite presentation, so is $\mathop{\mathrm{Coker}}(g)$. Then by Lemma 10.88.5, $f_ i : M_ i \to M$ factors through $g: M_ i \to Q$, say $f_ i = h \circ g$ for some $h: Q \to M$. Then since $Q$ is finitely presented, $h$ factors through $M_ j \to M$ for some $j \geq i$, say $h = f_ j \circ h'$ for some $h': Q \to M_ j$. In total we have a commutative diagram

$\xymatrix{ & M & \\ M_ i \ar[dr]_ g \ar[ur]^{f_ i} \ar[rr]^{f_{ij}} & & M_ j \ar[ul]_{f_ j} \\ & Q \ar[ur]_{h'} & }$

Thus $f_{ij}$ dominates $g$. But $g$ dominates $f_ i$, so $f_{ij}$ dominates $f_ i$.

Conversely, suppose (2) holds. Let $P$ be of finite presentation and $f: P \to M$ a module map. Then $f$ factors through $f_ i: M_ i \to M$ for some $i \in I$, say $f = f_ i \circ g'$ for some $g': P \to M_ i$. Choose by (2) a $j \geq i$ such that $f_{ij}$ dominates $f_ i$. We have a commutative diagram

$\xymatrix{ P \ar[d]_{g'} \ar[r]^{f} & M \\ M_ i \ar[ur]^{f_ i} \ar[r]_{f_{ij}} & M_ j \ar[u]_{f_ j} }$

From the diagram and the fact that $f_{ij}$ dominates $f_ i$, we find that $f$ and $f_{ij} \circ g'$ dominate each other. Hence taking $g = f_{ij} \circ g' : P \to M_ j$ works.

Next we prove (2) is equivalent to (3). Let $i \in I$. It is always true that $f_ i$ dominates $f_{ik}$ for $k \geq i$, since $f_ i$ factors through $f_{ik}$. If (2) holds, choose $j \geq i$ such that $f_{ij}$ dominates $f_ i$. Then since domination is a transitive relation, $f_{ij}$ dominates $f_{ik}$ for $k \geq i$. All $M_ i$ are of finite presentation, so $\mathop{\mathrm{Coker}}(f_{ik})$ is of finite presentation for $k \geq i$. By Lemma 10.88.5, $f_{ij}$ factors through $f_{ik}$ for all $k \geq i$. Thus (2) implies (3). On the other hand, if (3) holds then for any $R$-module $N$, $f_{ij} \otimes _ R \text{id}_ N$ factors through $f_{ik} \otimes _ R \text{id}_ N$ for $k \geq i$. So $\mathop{\mathrm{Ker}}(f_{ik} \otimes _ R \text{id}_ N) \subset \mathop{\mathrm{Ker}}(f_{ij} \otimes _ R \text{id}_ N)$ for $k \geq i$. But $\mathop{\mathrm{Ker}}(f_ i \otimes _ R \text{id}_ N: M_ i \otimes _ R N \to M \otimes _ R N)$ is the union of $\mathop{\mathrm{Ker}}(f_{ik} \otimes _ R \text{id}_ N)$ for $k \geq i$. Thus $\mathop{\mathrm{Ker}}(f_ i \otimes _ R \text{id}_ N) \subset \mathop{\mathrm{Ker}}(f_{ij} \otimes _ R \text{id}_ N)$ for any $R$-module $N$, which by definition means $f_{ij}$ dominates $f_ i$.

It is trivial that (3) implies (4) implies (5). We show (5) implies (3). Let $N = \prod _{s \in I} M_ s$. If (5) holds, then given $i \in I$ choose $j \geq i$ such that

$\mathop{\mathrm{Im}}( \mathop{\mathrm{Hom}}\nolimits (M_ j, N) \to \mathop{\mathrm{Hom}}\nolimits (M_ i, N)) = \mathop{\mathrm{Im}}( \mathop{\mathrm{Hom}}\nolimits (M_ k, N) \to \mathop{\mathrm{Hom}}\nolimits (M_ i, N))$

for all $k \geq j$. Passing the product over $s \in I$ outside of the $\mathop{\mathrm{Hom}}\nolimits$'s and looking at the maps on each component of the product, this says

$\mathop{\mathrm{Im}}( \mathop{\mathrm{Hom}}\nolimits (M_ j, M_ s) \to \mathop{\mathrm{Hom}}\nolimits (M_ i, M_ s)) = \mathop{\mathrm{Im}}( \mathop{\mathrm{Hom}}\nolimits (M_ k, M_ s) \to \mathop{\mathrm{Hom}}\nolimits (M_ i, M_ s))$

for all $k \geq j$ and $s \in I$. Taking $s = j$ we have

$\mathop{\mathrm{Im}}( \mathop{\mathrm{Hom}}\nolimits (M_ j, M_ j) \to \mathop{\mathrm{Hom}}\nolimits (M_ i, M_ j)) = \mathop{\mathrm{Im}}( \mathop{\mathrm{Hom}}\nolimits (M_ k, M_ j) \to \mathop{\mathrm{Hom}}\nolimits (M_ i, M_ j))$

for all $k \geq j$. Since $f_{ij}$ is the image of $\text{id} \in \mathop{\mathrm{Hom}}\nolimits (M_ j, M_ j)$ under $\mathop{\mathrm{Hom}}\nolimits (M_ j, M_ j) \to \mathop{\mathrm{Hom}}\nolimits (M_ i, M_ j)$, this shows that for any $k \geq j$ there is $h \in \mathop{\mathrm{Hom}}\nolimits (M_ k, M_ j)$ such that $f_{ij} = h \circ f_{ik}$. If $j \geq k$ then we can take $h = f_{kj}$. Hence (3) holds. $\square$

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