** A functor with an exact left adjoint preserves injectives **

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.
$\square$

## Comments (1)

Comment #5064 by Remy on

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