Lemma 10.89.7. Let $0 \to M_1 \to M_2 \to M_3 \to 0$ be a universally exact sequence of $R$-modules. Then:

If $M_2$ is Mittag-Leffler, then $M_1$ is Mittag-Leffler.

If $M_1$ and $M_3$ are Mittag-Leffler, then $M_2$ is Mittag-Leffler.

Lemma 10.89.7. Let $0 \to M_1 \to M_2 \to M_3 \to 0$ be a universally exact sequence of $R$-modules. Then:

If $M_2$ is Mittag-Leffler, then $M_1$ is Mittag-Leffler.

If $M_1$ and $M_3$ are Mittag-Leffler, then $M_2$ is Mittag-Leffler.

**Proof.**
For any family $(Q_{\alpha })_{\alpha \in A}$ of $R$-modules we have a commutative diagram

\[ \xymatrix{ 0 \ar[r] & M_1 \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d] & M_2 \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d] & M_3 \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \prod _{\alpha }(M_1 \otimes Q_{\alpha }) \ar[r] & \prod _{\alpha }(M_2 \otimes Q_{\alpha }) \ar[r] & \prod _{\alpha }(M_3 \otimes Q_{\alpha })\ar[r] & 0 } \]

with exact rows. Thus (1) and (2) follow from Proposition 10.89.5. $\square$

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (4)

Comment #2973 by Fred Rohrer on

Comment #2974 by Fred Rohrer on

Comment #2975 by Fred Rohrer on

Comment #3098 by Johan on

There are also: