Let $F: \mathcal{A} \longrightarrow \mathcal{B}$, $G: \mathcal{B} \longrightarrow \mathcal{A}$ be two additive functors between abelian categories, such that $(F, G)$ is an adjoint pair. I want to prove that, if there exist a cogenerator $B \in \mathcal{B}$ such that $G(B) \in \mathcal{A}$ is injective, then $F $ is exact.
I tried to prove this taking a short exact sequence $0 \longrightarrow A \longrightarrow A'\longrightarrow A'' \longrightarrow 0$ in $\mathcal{A}$, with the purpose to prove that also $0 \longrightarrow F(A) \longrightarrow F(A')\longrightarrow F(A'') \longrightarrow 0$ is exact in $\mathcal{B}$.
If $G(B)$ is injective in $\mathcal{A}$, then the functor $Hom_{\mathcal{A}}(-,G(B))$ is exact. Then, the sequence $0 \longrightarrow Hom_{\mathcal{A}}(A'',G(B)) \longrightarrow Hom_{\mathcal{A}}(A',G(B))\longrightarrow Hom_{\mathcal{A}}(A,G(B)) \longrightarrow 0$ is exact. In fact, by adjointness, the first of this sets is isomorphic to $Hom_{\mathcal{B}}(F(A''),B)$, and analogously the others.
How can I step on from this point?
Since $B$ is a cogenerator, $0 \to F A \to F A'$ is exact if $\mathrm{Hom} (F A', B) \to \mathrm{Hom} (F A, B) \to 0$ is exact. Since $F$ is a left adjoint, we already know that $F A \to F A' \to F A'' \to 0$ is exact, and as you say, $\mathrm{Hom} (F A', B) \to \mathrm{Hom} (F A, B) \to 0$ is exact, so we are done.