Lemma 10.92.2. Let $R$ be a ring. Let $M$ be an $R$-module. Assume $M$ is Mittag-Leffler and countably generated. For any $R$-module map $f : P \to M$ with $P$ finitely generated there exists an endomorphism $\alpha : M \to M$ such that

1. $\alpha : M \to M$ factors through a finitely presented $R$-module, and

2. $\alpha \circ f = f$.

Proof. Write $M = \mathop{\mathrm{colim}}\nolimits _{i \in I} M_ i$ as a directed colimit of finitely presented $R$-modules with $I$ countable, see Lemma 10.92.1. The transition maps are denoted $f_{ij}$ and we use $f_ i : M_ i \to M$ to denote the canonical maps into $M$. Set $N = \prod _{s \in I} M_ s$. Denote

$M_ i^* = \mathop{\mathrm{Hom}}\nolimits _ R(M_ i, N) = \prod \nolimits _{s \in I} \mathop{\mathrm{Hom}}\nolimits _ R(M_ i, M_ s)$

so that $(M_ i^*)$ is an inverse system of $R$-modules over $I$. Note that $\mathop{\mathrm{Hom}}\nolimits _ R(M, N) = \mathop{\mathrm{lim}}\nolimits M_ i^*$. As $M$ is Mittag-Leffler, we find for every $i \in I$ an index $k(i) \geq i$ such that

$E_ i := \bigcap \nolimits _{i' \geq i} \mathop{\mathrm{Im}}(M_{i'}^* \to M_ i^*) = \mathop{\mathrm{Im}}(M_{k(i)}^* \to M_ i^*)$

Choose and fix $j \in I$ such that $\mathop{\mathrm{Im}}(P \to M) \subset \mathop{\mathrm{Im}}(M_ j \to M)$. This is possible as $P$ is finitely generated. Set $k = k(j)$. Let $x = (0, \ldots , 0, \text{id}_{M_ k}, 0, \ldots , 0) \in M_ k^*$ and note that this maps to $y = (0, \ldots , 0, f_{jk}, 0, \ldots , 0) \in M_ j^*$. By our choice of $k$ we see that $y \in E_ j$. By Example 10.86.2 the transition maps $E_ i \to E_ j$ are surjective for each $i \geq j$ and $\mathop{\mathrm{lim}}\nolimits E_ i = \mathop{\mathrm{lim}}\nolimits M_ i^* = \mathop{\mathrm{Hom}}\nolimits _ R(M, N)$. Hence Lemma 10.86.3 guarantees there exists an element $z \in \mathop{\mathrm{Hom}}\nolimits _ R(M, N)$ which maps to $y$ in $E_ j \subset M_ j^*$. Let $z_ k$ be the $k$th component of $z$. Then $z_ k : M \to M_ k$ is a homomorphism such that

$\xymatrix{ M \ar[r]_{z_ k} & M_ k \\ M_ j \ar[ru]_{f_{jk}} \ar[u]^{f_ j} }$

commutes. Let $\alpha : M \to M$ be the composition $f_ k \circ z_ k : M \to M_ k \to M$. Then $\alpha$ factors through a finitely presented module by construction and $\alpha \circ f_ j = f_ j$. Since the image of $f$ is contained in the image of $f_ j$ this also implies that $\alpha \circ f = f$. $\square$

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