## 12.29 Injectives and adjoint functors

Here are some lemmas on adjoint functors and their relationship with injectives. See also Lemma 12.7.4.

slogan
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 with $u$ right adjoint to $v$. Consider the following conditions:

$v$ transforms injective maps into injective maps,

$v$ is exact, and

$u$ transforms injectives into injectives.

Then (a) $\Leftrightarrow $ (b) $\Rightarrow $ (c). If $\mathcal{A}$ has enough injectives, then all three conditions are equivalent.

**Proof.**
Observe that $v$ is right exact as a left adjoint (Categories, Lemma 4.24.6). Combined with Lemma 12.7.2 this explains why (a) $\Leftrightarrow $ (b).

Assume (a). 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 $\beta : M \to uI$ extending $\alpha $. Hence (c) is true.

Assume $\mathcal{A}$ has enough injectives and (c) holds. Let $f : B \to B'$ be a monomorphism in $\mathcal{B}$, and let $A = \mathop{\mathrm{Ker}}(v(f))$. Choose a monomorphism $g : A \to I$ with $I$ injective (it exists by assumption). Then $g$ extends to $g' : v(B) \to I$, whence by adjunction a morphism $B \to u(I)$. Since $u(I)$ is injective, this morphism extends to $h : B' \to u(I)$, hence by adjunction a morphism $k : v(B') \to I$ extending $g'$. But then $k$ "extendsâ€ť $g$, which forces $A = 0$ since $g$ was a monomorphism. Thus (a) is true.
$\square$

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

$u$ is right adjoint to $v$,

$v$ transforms injective maps into injective maps,

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

$vB = 0$ 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.
$\square$

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

$u$ is right adjoint to $v$,

$v$ transforms injective maps into injective maps,

$\mathcal{A}$ has enough injectives,

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

$\mathcal{A}$ has functorial injective embeddings.

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

**Proof.**
Let $A \mapsto (A \to J(A))$ be a functorial injective embedding on $\mathcal{A}$. Then $B \mapsto (B \to uJ(vB))$ is a functorial injective embedding 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

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

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.
$\square$

## Comments (2)

Comment #8711 by Bruno Kahn on

Comment #9362 by Stacks project on