Injective Cogenerators in the Category of Modules over a Noetherian Ring

Let $R$ be a Noetherian ring and let $\mathcal{A}$ be an injective $R$-module. The injectivity of $\mathcal{A}$ is equivalent to the exactness of the functor $Hom_R(-,\mathcal{A})$, i.e. whenever we have an exact sequence of $R$-modules $M \rightarrow N \rightarrow P, \, \, \, (1)$, then the sequence $Hom_R(P,\mathcal{A}) \rightarrow Hom_R(N,\mathcal{A}) \rightarrow Hom_R(M,\mathcal{A}), \, \, \, (2)$ is also exact. Now assume that $\mathcal{A}$ is also a cogenerator in the category of $R$-modules, i.e. $Hom_R(M,\mathcal{A})=0 \Rightarrow M=0$. How can we prove that exactness of $(2)$ implies exactness of $(1)$?

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Let $\mathcal{C}$ be an abelian category, and suppose $A$ is an injective cogenerator in $\mathcal{C}$. For any morphism $f : X \to Y$ in $\mathcal{C}$, consider the following exact sequence in $\mathcal{C}$: $$0 \longrightarrow W \longrightarrow X \longrightarrow Y \longrightarrow Z \longrightarrow 0$$ Since $A$ is injective, we get an exact sequence of abelian groups: $$0 \longrightarrow \mathcal{C}(Z, A) \longrightarrow \mathcal{C}(Y, A) \longrightarrow \mathcal{C}(X, A) \longrightarrow \mathcal{C}(W, A) \longrightarrow 0$$ So, if $\mathcal{C}(f, A) : \mathcal{C}(Y, A) \to \mathcal{C}(X, A)$ is an isomorphism in $\textbf{Ab}$, then $\mathcal{C}(Z, A)$ and $\mathcal{C}(W, A)$ must be trivial, and this implies $Z$ and $W$ are zero in $\mathcal{C}$ because $A$ is a cogenerator. Thus, $\mathcal{C}(f, A) : \mathcal{C}(Y, A) \to \mathcal{C}(X, A)$ is an isomorphism if and only if $f : X \to Y$ is an isomorphism in $\mathcal{C}$; in other words, $\mathcal{C}(-, A) : \mathcal{C}^\textrm{op} \to \textbf{Ab}$ is a conservative functor.
Now, it is a general fact that conservative functors reflect all limits and colimits that they preserve, so this implies $\mathcal{C}(-, A)$ reflects exact sequences. For example, suppose we have a chain complex in $\mathcal{C}$ $$X \longrightarrow Y \longrightarrow Z \longrightarrow 0$$ whose image in $\textbf{Ab}$ is exact: $$0 \longrightarrow \mathcal{C}(Z, A) \longrightarrow \mathcal{C}(Y, A) \longrightarrow \mathcal{C}(X, A)$$ Now compute the cokernel in $\mathcal{C}$ so that we get an exact sequence $$X \longrightarrow Y \longrightarrow C \longrightarrow 0$$ and the universal property of $C$ ensures there is a chain morphism from this complex to the first one; but because the image of the first complex is already exact, the image of the morphism $C \to Z$ must be an isomorphism in $\textbf{Ab}$, so $C \to Z$ itself must be an isomorphism in $\mathcal{C}$, and therefore the first complex must be exact in $\mathcal{C}$.
Thanks for your very interesting answer. When we write $Y \rightarrow Z \rightarrow 0$ in an abelian category, what exactly do we mean? Is it that the morphism $Y \rightarrow Z$ has zero cockernel ($Y \rightarrow Z$ is surjective)? But the rest of the chain complex is not exact, right? More precisely, what exactly do you mean by $X \rightarrow Y \rightarrow Z \rightarrow 0$ in an abelian category? (my knowledge of abelian categories is that of Lang's algebra p. 133). –  Manos Nov 6 '12 at 15:46
Additionally, in my context, i do not assume anything about the exactness of my sequence $M \rightarrow N \rightarrow P$. Then why do you consider a complex of the form $X \rightarrow Y \rightarrow Z \rightarrow 0$? –  Manos Nov 6 '12 at 15:52
$Y \to Z \to 0$ is exact if and only if $Y \to Z$ is an epimorphism. $X \to Y \to Z \to 0$ is exact if and only if $Y \to Z$ is the cokernel of $X \to Y$. I assume $X \to Y \to Z \to 0$ is a chain complex for simplicity: but actually you can show that $\mathcal{C}(f, A) = 0$ if and only if $f = 0$ when $A$ is an injective cogenerator. –  Zhen Lin Nov 6 '12 at 17:48