## 12.29 Injectives and adjoint functors

Lemma 12.29.1. Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories. Let $u : \mathcal{A} \to \mathcal{B}$ and $v : \mathcal{B} \to \mathcal{A}$ be additive functors. Assume

1. $u$ is right adjoint to $v$, and

2. $v$ transforms injective maps into injective maps.

Then $u$ transforms injectives into injectives.

Proof. Let $I$ be an injective object of $\mathcal{A}$. Let $\varphi : N \to M$ be an injective map in $\mathcal{B}$ and let $\alpha : N \to uI$ be a morphism. By adjointness we get a morphism $\alpha : vN \to I$ and by assumption $v\varphi : vN \to vM$ is injective. Hence as $I$ is an injective object we get a morphism $\beta : vM \to I$ extending $\alpha$. By adjointness again this corresponds to a morphism $\beta : M \to uI$ as desired. $\square$

Remark 12.29.2. Let $\mathcal{A}$, $\mathcal{B}$, $u : \mathcal{A} \to \mathcal{B}$ and $v : \mathcal{B} \to \mathcal{A}$ be as in Lemma 12.29.1. In the presence of assumption (1) assumption (2) is equivalent to requiring that $v$ is exact. Moreover, condition (2) is necessary. Here is an example. Let $A \to B$ be a ring map. Let $u : \text{Mod}_ B \to \text{Mod}_ A$ be $u(N) = N_ A$ and let $v : \text{Mod}_ A \to \text{Mod}_ B$ be $v(M) = M \otimes _ A B$. Then $u$ is right adjoint to $v$, and $u$ is exact and $v$ is right exact, but $v$ does not transform injective maps into injective maps in general (i.e., $v$ is not left exact). Moreover, it is not the case that $u$ transforms injective $B$-modules into injective $A$-modules. For example, if $A = \mathbf{Z}$ and $B = \mathbf{Z}/p\mathbf{Z}$, then the injective $B$-module $\mathbf{Z}/p\mathbf{Z}$ is not an injective $\mathbf{Z}$-module. In fact, the lemma applies to this example if and only if the ring map $A \to B$ is flat.

Lemma 12.29.3. Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories. Let $u : \mathcal{A} \to \mathcal{B}$ and $v : \mathcal{B} \to \mathcal{A}$ be additive functors. Assume

1. $u$ is right adjoint to $v$,

2. $v$ transforms injective maps into injective maps,

3. $\mathcal{A}$ has enough injectives, and

4. $vB = 0$ implies $B = 0$ for any $B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$.

Then $\mathcal{B}$ has enough injectives.

Proof. Pick $B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$. Pick an injection $vB \to I$ for $I$ an injective object of $\mathcal{A}$. According to Lemma 12.29.1 and the assumptions the corresponding map $B \to uI$ is the injection of $B$ into an injective object. $\square$

Remark 12.29.4. Let $\mathcal{A}$, $\mathcal{B}$, $u : \mathcal{A} \to \mathcal{B}$ and $v : \mathcal{B} \to \mathcal{A}$ be as In Lemma 12.29.3. In the presence of conditions (1) and (2) condition (4) is equivalent to $v$ being faithful. Moreover, condition (4) is needed. An example is to consider the case where the functors $u$ and $v$ are both the zero functor.

Lemma 12.29.5. Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories. Let $u : \mathcal{A} \to \mathcal{B}$ and $v : \mathcal{B} \to \mathcal{A}$ be additive functors. Assume

1. $u$ is right adjoint to $v$,

2. $v$ transforms injective maps into injective maps,

3. $\mathcal{A}$ has enough injectives,

4. $vB = 0$ implies $B = 0$ for any $B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$, and

5. $\mathcal{A}$ has functorial injective hulls.

Then $\mathcal{B}$ has functorial injective hulls.

Proof. Let $A \mapsto (A \to J(A))$ be a functorial injective hull on $\mathcal{A}$. Then $B \mapsto (B \to uJ(vB))$ is a functorial injective hull on $\mathcal{B}$. Compare with the proof of Lemma 12.29.3. $\square$

Lemma 12.29.6. Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories. Let $u : \mathcal{A} \to \mathcal{B}$ be a functor. If there exists a subset $\mathcal{P} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$ such that

1. every object of $\mathcal{B}$ is a quotient of an element of $\mathcal{P}$, and

2. for every $P \in \mathcal{P}$ there exists an object $Q$ of $\mathcal{A}$ such that $\mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(Q, A) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(P, u(A))$ functorially in $A$,

then there exists a left adjoint $v$ of $u$.

Proof. By the Yoneda lemma (Categories, Lemma 4.3.5) the object $Q$ of $\mathcal{A}$ corresponding to $P$ is defined up to unique isomorphism by the formula $\mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(Q, A) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(P, u(A))$. Let us write $Q = v(P)$. Denote $i_ P : P \to u(v(P))$ the map corresponding to $\text{id}_{v(P)}$ in $\mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(v(P), v(P))$. Functoriality in (2) implies that the bijection is given by

$\mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(v(P), A) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(P, u(A)),\quad \varphi \mapsto u(\varphi ) \circ i_ P$

For any pair of elements $P_1, P_2 \in \mathcal{P}$ there is a canonical map

$\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(P_2, P_1) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(v(P_2), v(P_1)),\quad \varphi \mapsto v(\varphi )$

which is characterized by the rule $u(v(\varphi )) \circ i_{P_2} = i_{P_1} \circ \varphi$ in $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(P_2, u(v(P_1)))$. Note that $\varphi \mapsto v(\varphi )$ is compatible with composition; this can be seen directly from the characterization. Hence $P \mapsto v(P)$ is a functor from the full subcategory of $\mathcal{B}$ whose objects are the elements of $\mathcal{P}$.

Given an arbitrary object $B$ of $\mathcal{B}$ choose an exact sequence

$P_2 \to P_1 \to B \to 0$

which is possible by assumption (1). Define $v(B)$ to be the object of $\mathcal{A}$ fitting into the exact sequence

$v(P_2) \to v(P_1) \to v(B) \to 0$

Then

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(v(B), A) & = \mathop{\mathrm{Ker}}(\mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(v(P_1), A) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(v(P_2), A)) \\ & = \mathop{\mathrm{Ker}}(\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(P_1, u(A)) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(P_2, u(A))) \\ & = \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(B, u(A)) \end{align*}

Hence we see that we may take $\mathcal{P} = \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$, i.e., we see that $v$ is everywhere defined. $\square$

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