The Stacks project

Lemma 35.4.12. Let $R$ be a ring. A morphism $f: M \to N$ in $\text{Mod}_ R$ is universally injective if and only if $C(f): C(N) \to C(M)$ is a split surjection.

Proof. By (, for any $P \in \text{Mod}_ R$ we have a commutative diagram

\[ \xymatrix@C=9pc{ \mathop{\mathrm{Hom}}\nolimits _ R( P, C(N)) \ar[r]_{\mathop{\mathrm{Hom}}\nolimits _ R(P,C(f))} \ar[d]^{\cong } & \mathop{\mathrm{Hom}}\nolimits _ R(P,C(M)) \ar[d]^{\cong } \\ C(P \otimes _ R N ) \ar[r]^{C(1_{P} \otimes f)} & C(P \otimes _ R M ). } \]

If $f$ is universally injective, then $1_{C(M)} \otimes f: C(M) \otimes _ R M \to C(M) \otimes _ R N$ is injective, so both rows in the above diagram are surjective for $P = C(M)$. We may thus lift $1_{C(M)} \in \mathop{\mathrm{Hom}}\nolimits _ R(C(M), C(M))$ to some $g \in \mathop{\mathrm{Hom}}\nolimits _ R(C(N), C(M))$ splitting $C(f)$. Conversely, if $C(f)$ is a split surjection, then both rows in the above diagram are surjective, so by Lemma 35.4.10, $1_{P} \otimes f$ is injective. $\square$

Comments (0)

There are also:

  • 4 comment(s) on Section 35.4: Descent for universally injective morphisms

Post a comment

Your email address will not be published.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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 08WU. Beware of the difference between the letter 'O' and the digit '0'.