# The Stacks Project

## Tag 08WJ

#### 34.4.3. 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.

Remark 34.4.4. 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.

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 10.81.1.

Definition 34.4.5. A ring map $f: R \to S$ is universally injective if it is universally injective as a morphism in $\text{Mod}_R$.

Example 34.4.6. Any split injection in $\text{Mod}_R$ is universally injective. In particular, any split injection in $\textit{Rings}$ is universally injective.

Example 34.4.7. 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 34.4.8, 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 34.4.8. Any faithfully flat ring map is universally injective.

Proof. This is a reformulation of Algebra, Lemma 10.81.11. $\square$

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 34.4.9. Let $R$ be a ring. Define the contravariant functor $C$ $: \text{Mod}_R \to \text{Mod}_R$ by setting $$C(M) = \mathop{\rm Hom}\nolimits_{\textit{Ab}}(M, \mathbf{Q}/\mathbf{Z}),$$ 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 15.51.

Lemma 34.4.10. 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 15.51.6 and the other properties follow from this, see Remark 34.4.4. $\square$

Remark 34.4.11. We will use frequently the standard adjunction between $\mathop{\rm Hom}\nolimits$ and tensor product, in the form of the natural isomorphism of contravariant functors $$\tag{34.4.11.1} C(\bullet_1 \otimes_R \bullet_2) \cong \mathop{\rm Hom}\nolimits_R(\bullet_1, C(\bullet_2)): \text{Mod}_R \times \text{Mod}_R \to \text{Mod}_R$$ 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 10.13.5. A corollary of this observation is that if $$\xymatrix@C=9pc{ C(M) \ar@<1ex>[r] \ar@<-1ex>[r] & C(N) \ar[r] & C(P) }$$ is a split coequalizer diagram in $\text{Mod}_R$, then so is $$\xymatrix@C=9pc{ C(M \otimes_R Q) \ar@<1ex>[r] \ar@<-1ex>[r] & C(N \otimes_R Q) \ar[r] & C(P \otimes_R Q) }$$ for any $Q \in \text{Mod}_R$.

Lemma 34.4.12. 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 (34.4.11.1), for any $P \in \text{Mod}_R$ we have a commutative diagram $$\xymatrix@C=9pc{ \mathop{\rm Hom}\nolimits_R( P, C(N)) \ar[r]_{\mathop{\rm Hom}\nolimits_R(P,C(f))} \ar[d]^{\cong} & \mathop{\rm Hom}\nolimits_R(P,C(M)) \ar[d]^{\cong} \\ C(P \otimes_R N ) \ar[r]^{C(1_{P} \otimes f)} & C(P \otimes_R M ). }$$ 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 $1_{C(M)} \in \mathop{\rm Hom}\nolimits_R(C(M), C(M))$ to some $g \in \mathop{\rm Hom}\nolimits_R(C(N), C(M))$ splitting $C(f)$. Conversely, if $C(f)$ is a split surjection, then both rows in the above diagram are surjective, so by Lemma 34.4.10, $1_{P} \otimes f$ is injective. $\square$

Remark 34.4.13. 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 (34.4.11.1) this amounts to splitting $\mathop{\rm Hom}\nolimits_R(P, C(f))$ functorially in $P$, and this is achieved by the map $g \circ \bullet$.

The code snippet corresponding to this tag is a part of the file descent.tex and is located in lines 878–1034 (see updates for more information).

\subsection{Universally injective morphisms}
\label{subsection-universally-injective}

\noindent
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.

\begin{remark}
\label{remark-reflects}
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 \ref{homology-section-functors}).
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.
\end{remark}

\noindent
Let $R$ be a ring. Recall that a morphism $f : M \to N$ in $\text{Mod}_R$
is {\it 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 \ref{algebra-definition-universally-injective}.

\begin{definition}
\label{definition-universally-injective}
A ring map $f: R \to S$ is {\it universally injective}
if it is universally injective as a morphism in $\text{Mod}_R$.
\end{definition}

\begin{example}
\label{example-split-injection-universally-injective}
Any split injection in $\text{Mod}_R$ is universally injective. In particular,
any split injection in $\textit{Rings}$ is universally injective.
\end{example}

\begin{example}
\label{example-cover-universally-injective}
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 \ref{lemma-faithfully-flat-universally-injective},
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.
\end{example}

\begin{lemma}
\label{lemma-faithfully-flat-universally-injective}
Any faithfully flat ring map is universally injective.
\end{lemma}

\begin{proof}
This is a reformulation of Algebra, Lemma
\ref{algebra-lemma-faithfully-flat-universally-injective}.
\end{proof}

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

\begin{definition}
\label{definition-C}
Let $R$ be a ring. Define the contravariant functor
{\it $C$} $: \text{Mod}_R \to \text{Mod}_R$ by setting
$$C(M) = \Hom_{\textit{Ab}}(M, \mathbf{Q}/\mathbf{Z}),$$
with the $R$-action on $C(M)$ given by $rf(s) = f(rs)$.
\end{definition}

\noindent
This functor was denoted $M \mapsto M^\vee$ in
More on Algebra, Section \ref{more-algebra-section-injectives-modules}.

\begin{lemma}
\label{lemma-C-is-faithful}
For a ring $R$, the functor $C : \text{Mod}_R \to \text{Mod}_R$ is
exact and reflects injections and surjections.
\end{lemma}

\begin{proof}
Exactness is More on Algebra, Lemma \ref{more-algebra-lemma-vee-exact}
and the other properties follow from this, see
Remark \ref{remark-reflects}.
\end{proof}

\begin{remark}
We will use frequently the standard adjunction between $\Hom$ and tensor
product, in the form of the natural isomorphism of contravariant functors

C(\bullet_1 \otimes_R \bullet_2) \cong \Hom_R(\bullet_1, C(\bullet_2)):
\text{Mod}_R \times \text{Mod}_R \to \text{Mod}_R

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 \ref{algebra-lemma-hom-from-tensor-product-variant}.
A corollary of this observation is that if
$$\xymatrix@C=9pc{ C(M) \ar@<1ex>[r] \ar@<-1ex>[r] & C(N) \ar[r] & C(P) }$$
is a split coequalizer diagram in $\text{Mod}_R$, then so is
$$\xymatrix@C=9pc{ C(M \otimes_R Q) \ar@<1ex>[r] \ar@<-1ex>[r] & C(N \otimes_R Q) \ar[r] & C(P \otimes_R Q) }$$
for any $Q \in \text{Mod}_R$.
\end{remark}

\begin{lemma}
\label{lemma-split-surjection}
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.
\end{lemma}

\begin{proof}
By (\ref{equation-adjunction}), for any $P \in \text{Mod}_R$ we have a
commutative diagram
$$\xymatrix@C=9pc{ \Hom_R( P, C(N)) \ar[r]_{\Hom_R(P,C(f))} \ar[d]^{\cong} & \Hom_R(P,C(M)) \ar[d]^{\cong} \\ C(P \otimes_R N ) \ar[r]^{C(1_{P} \otimes f)} & C(P \otimes_R M ). }$$
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
$1_{C(M)} \in \Hom_R(C(M), C(M))$ to some $g \in \Hom_R(C(N), C(M))$ splitting
$C(f)$.
Conversely, if $C(f)$ is a split surjection, then
both rows in the above diagram are surjective,
so by Lemma \ref{lemma-C-is-faithful}, $1_{P} \otimes f$ is injective.
\end{proof}

\begin{remark}
\label{remark-functorial-splitting}
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 (\ref{equation-adjunction}) this amounts to splitting $\Hom_R(P, C(f))$  functorially in $P$,
and this is achieved by the map $g \circ \bullet$.
\end{remark}

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