Remark 35.4.4. Any functor $F : \mathcal{A} \to \mathcal{B}$ of abelian categories which is exact and takes nonzero objects to nonzero objects reflects injections and surjections. Namely, exactness implies that $F$ preserves kernels and cokernels (compare with Homology, Section 12.7). For example, if $f : R \to S$ is a faithfully flat ring homomorphism, then $\bullet \otimes _ R S: \text{Mod}_ R \to \text{Mod}_ S$ has these properties.

### 35.4.3 Universally injective morphisms

Recall that $\textit{Rings}$ denotes the category of commutative rings with $1$. For an object $R$ of $\textit{Rings}$ we denote $\text{Mod}_ R$ the category of $R$-modules.

Let $R$ be a ring. Recall that a morphism $f : M \to N$ in $\text{Mod}_ R$ is *universally injective* if for all $P \in \text{Mod}_ R$, the morphism $f \otimes 1_ P: M \otimes _ R P \to N \otimes _ R P$ is injective. See Algebra, Definition 10.82.1.

Definition 35.4.5. A ring map $f: R \to S$ is *universally injective* if it is universally injective as a morphism in $\text{Mod}_ R$.

Example 35.4.6. Any split injection in $\text{Mod}_ R$ is universally injective. In particular, any split injection in $\textit{Rings}$ is universally injective.

Example 35.4.7. For a ring $R$ and $f_1, \ldots , f_ n \in R$ generating the unit ideal, the morphism $R \to R_{f_1} \oplus \ldots \oplus R_{f_ n}$ is universally injective. Although this is immediate from Lemma 35.4.8, it is instructive to check it directly: we immediately reduce to the case where $R$ is local, in which case some $f_ i$ must be a unit and so the map $R \to R_{f_ i}$ is an isomorphism.

Lemma 35.4.8. Any faithfully flat ring map is universally injective.

**Proof.**
This is a reformulation of Algebra, Lemma 10.82.11.
$\square$

The key observation from [mesablishvili1] is that universal injectivity can be usefully reformulated in terms of a splitting, using the usual construction of an injective cogenerator in $\text{Mod}_ R$.

Definition 35.4.9. Let $R$ be a ring. Define the contravariant functor *$C$* $ : \text{Mod}_ R \to \text{Mod}_ R$ by setting

with the $R$-action on $C(M)$ given by $rf(s) = f(rs)$.

This functor was denoted $M \mapsto M^\vee $ in More on Algebra, Section 15.55.

Lemma 35.4.10. For a ring $R$, the functor $C : \text{Mod}_ R \to \text{Mod}_ R$ is exact and reflects injections and surjections.

**Proof.**
Exactness is More on Algebra, Lemma 15.55.6 and the other properties follow from this, see Remark 35.4.4.
$\square$

Remark 35.4.11. We will use frequently the standard adjunction between $\mathop{\mathrm{Hom}}\nolimits $ and tensor product, in the form of the natural isomorphism of contravariant functors

taking $f: M_1 \otimes _ R M_2 \to \mathbf{Q}/\mathbf{Z}$ to the map $m_1 \mapsto (m_2 \mapsto f(m_1 \otimes m_2))$. See Algebra, Lemma 10.14.5. A corollary of this observation is that if

is a split coequalizer diagram in $\text{Mod}_ R$, then so is

for any $Q \in \text{Mod}_ R$.

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

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. $\square$

Remark 35.4.13. Let $f: M \to N$ be a universally injective morphism in $\text{Mod}_ R$. By choosing a splitting $g$ of $C(f)$, we may construct a functorial splitting of $C(1_ P \otimes f)$ for each $P \in \text{Mod}_ R$. Namely, by (35.4.11.1) this amounts to splitting $\mathop{\mathrm{Hom}}\nolimits _ R(P, C(f))$ functorially in $P$, and this is achieved by the map $g \circ \bullet $.

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