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 (35.4.11.1), 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. $\square$
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