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Let $F:\mathcal{A} \rightarrow \mathcal{B}$ and $G:\mathcal{B} \rightarrow \mathcal{A}$ be additive functors between abelian categories, such that $(F,G)$ is an adjoint pair. If $B \in \mathcal{B}$ is a cogenerator and $G(B) \in \mathcal{A}$ is injective, then $F$ is exact.

I'm sorry if I'm asking something I've already found here Exactness of a right adjoint functor, but its argument does not convince me. All I can say is that:

I just need to prove that $F$ is left exact (I know $F$ is right exact by adjointness). Now I know $$Hom_\mathcal{A}(-,G(B)) \cong Hom_{\mathcal{B}}(F(-), B) = Hom_\mathcal{B}(-,B) \circ F $$ so that $Hom_{\mathcal{B}}(F(-), B)$ is exact by adjointness and $Hom_\mathcal{B}(-,B)$ is faithful since $B$ is a cogenerator. If $0\rightarrow A \rightarrow A'$ is exact, how can I deduce $0 \rightarrow F(A) \rightarrow F(A')$ is exact? Thanks for helping: I've spent hours on this exercise but I've kept missing something...

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The claim follows from this fact:

If $H$ is a faithful contravariant functor and $H f$ is an epimorphism, then $f$ is a monomorphism.

As you say, $\mathrm{Hom}(-, B)$ is faithful, so if $\mathrm{Hom} (F A', B) \to \mathrm{Hom} (F A, B)$ is an epimorphism, then $F A \to F A'$ is a monomorphism. But $\mathrm{Hom} (F {-}, B)$ is exact and $A \to A'$ is a monomorphism, so $F A \to F A'$ is indeed a monomorphism.

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Let $0 \to X \to Y$ be a sequence in $\mathcal B$ such that

$$\operatorname{Hom}(Y,B) \to \operatorname{Hom}(X,B) \to 0$$

is exact.

Now let $T$ be the kernel of $X \to Y$. The composition $T \to X \to Y$ is zero, hence the composition

$$\operatorname{Hom}(Y,B) \to \operatorname{Hom}(X,B) \to \operatorname{Hom}(T,B)$$ is zero.

By the surjectivity of the first map, we get that $\operatorname{Hom}(X,B) \to \operatorname{Hom}(T,B)$ is zero. By the faithfulness of $\operatorname{Hom}(-,B)$, we get that $T \to X$ is zero, i.e. $0 \to X \to Y$ is exact.

Use this with $X=F(A), Y=F(A')$.

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