Prove that if $\zeta$ a torsion theory cogenerated by $U$, then $\zeta=(\ker \text{Hom}_R(-,U),\text{Cog}(U))$

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Let $\zeta$ be a torsion theory given by $\zeta=(\mathcal{T},\mathcal{F})$ where $\mathcal{T}$ is a torsion class and $\mathcal{F}$ is the torsion-free class. Prove that if $\zeta$ is cogenerated by $U$, $$\mathcal{T}:=\{M\in R\text{-Mod}:\text{Hom}_R(M,U)=0\}$$ and $$\mathcal{F}:=\{N\in R\text{-Mod}:\text{Hom}_R(M,N)=0~\text{for all}~M\in \mathcal{T}\},$$ then $$\zeta=(\ker \text{Hom}_R(-,U),\text{Cog}(U)).$$Here $\text{Cog}(U))$ is class of R-modules finitely cogenerated by $U$.

Proof: I know that an object $C$ is a cogenerator of a category $\mathcal{C}$ if $\text{Hom}(-,C): C \to$ Sets is a faithful functor.

In the prove I already know that $\mathcal{T}:=\{M\in R\text{-Mod}:\text{Hom}_R(M,U)=0\}$ implies $$\mathcal{T}:=\ker\text{Hom}_R(-,U).$$ But I cannot see why $\mathcal{F}=\text{Cog}(U)$?