Lemma 10.88.10. Let $R$ be a ring and $M$ an $R$-module. Then $M$ is Mittag-Leffler if and only if for every finite free $R$-module $F$ and module map $f: F \to M$, there exists a finitely presented $R$-module $Q$ and a module map $g : F \to Q$ 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 $N$.

Proof. Since the condition is clear weaker than condition (1) of Proposition 10.88.6 we see that a Mittag-Leffler module satisfies the condition. Conversely, suppose that $M$ satisfies the condition and that $f : P \to M$ is an $R$-module map from a finitely presented $R$-module $P$ into $M$. Choose a surjection $F \to P$ where $F$ is a finite free $R$-module. By assumption we can find a map $F \to Q$ where $Q$ is a finitely presented $R$-module such that $F \to Q$ and $F \to M$ dominate each other. In particular, the kernel of $F \to Q$ contains the kernel of $F \to P$, hence we obtain an $R$-module map $g : P \to Q$ such that $F \to Q$ is equal to the composition $F \to P \to Q$. Let $N$ be any $R$-module and consider the commutative diagram

$\xymatrix{ F \otimes _ R N \ar[d] \ar[r] & Q \otimes _ R N \\ P \otimes _ R N \ar[ru] \ar[r] & M \otimes _ R N }$

By assumption the kernels of $F \otimes _ R N \to Q \otimes _ R N$ and $F \otimes _ R N \to M \otimes _ R N$ are equal. Hence, as $F \otimes _ R N \to P \otimes _ R N$ is surjective, also the kernels of $P \otimes _ R N \to Q \otimes _ R N$ and $P \otimes _ R N \to M \otimes _ R N$ are equal. $\square$

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