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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